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Problem 25. Let \( a, b, c, d \) be positive real numbers such that \( {abcd} = 1 \) . Prove that\n\n\[ \frac{1}{{\left( 1 + a\right) }^{2}} + \frac{1}{{\left( 1 + b\right) }^{2}} + \frac{1}{{\left( 1 + c\right) }^{2}} + \frac{1}{{\left( 1 + d\right) }^{2}} \geq 1. \] | Solution. First, notice that for any non-negative real numbers \( x, y \)\n\n\[ \frac{1}{{\left( 1 + x\right) }^{2}} + \frac{1}{{\left( 1 + y\right) }^{2}} \geq \frac{1}{1 + {xy}} \]\n\nBy expanding, the above inequality becomes\n\n\[ \left( {2 + {2x} + {2y} + {x}^{2} + {y}^{2}}\right) \left( {1 + {xy}}\right) \geq \le... | Yes |
Problem 26. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a + {2b}}{c + {2b}} + \frac{b + {2c}}{a + {2c}} + \frac{c + {2a}}{b + {2a}} \geq 3. \] | Solution. By expanding, we can rewrite the inequality to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {a + {2b}}\right) \left( {a + {2c}}\right) \left( {b + {2a}}\right) \geq 3\mathop{\prod }\limits_{{cyc}}\left( {c + {2b}}\right) \]\n\n\[ \Leftrightarrow 2\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) + {3abc} \geq 3\left( ... | Yes |
Let \( a, b, c \) be three positive real numbers. Prove that\n\n\[ \frac{{a}^{4}}{{a}^{2} + {ab} + {b}^{2}} + \frac{{b}^{4}}{{b}^{2} + {bc} + {c}^{2}} + \frac{{c}^{4}}{{c}^{2} + {ca} + {a}^{2}} \geq \frac{{a}^{3} + {b}^{3} + {c}^{3}}{a + b + c}. \]\n\n(Phan Thanh Viet) | Solution. Notice that\n\n\[ \frac{{a}^{3} + {b}^{3} + {c}^{3}}{a + b + c} = \frac{3abc}{a + b + c} + {a}^{2} + {b}^{2} + {c}^{2} - {ab} - {bc} - {ca}. \]\n\nTherefore, the inequality can be rewritten in the following form\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{{a}^{4}}{{a}^{2} + {ab} + {b}^{2}} - {a}^{2} + {a... | Yes |
Problem 28. Prove that for all non-negative real numbers \( a, b, c \)\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + {2abc} + 1 \geq 2\left( {{ab} + {bc} + {ca}}\right) .\n\]\n\n(Darij Grinberg) | Solution. I will give four solutions to the above inequality.\n\nFirst Solution. Transform the inequality into a quadratic form in \( a \)\n\n\[ \nf\left( a\right) = {a}^{2} + 2\left( {{bc} - b - c}\right) a + {\left( b - c\right) }^{2} + 1.\n\]\n\n(i). If \( {bc} - b - c \geq 0 \), we get the desired result immediatel... | Yes |
Problem 29. Let \( a, b, c \) be positive real numbers and \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \frac{1}{2 - a} + \frac{1}{2 - b} + \frac{1}{2 - c} \geq 3 \]\n(Pham Kim Hung) | Solution.\n\nFirst Solution. (We use Cauchy-Schwarz inequality) Rewrite the inequality to\n\n\[ \left( {\frac{1}{2 - a} - \frac{1}{2}}\right) + \left( {\frac{1}{2 - b} - \frac{1}{2}}\right) + \left( {\frac{1}{2 - c} - \frac{1}{2}}\right) \geq \frac{3}{2} \]\n\n\[ \Leftrightarrow \frac{a}{2 - a} + \frac{b}{2 - b} + \fra... | Yes |
Problem 30. Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ \left( {1 + \frac{2a}{b + c}}\right) \left( {1 + \frac{2b}{c + d}}\right) \left( {1 + \frac{2c}{d + a}}\right) \left( {1 + \frac{2d}{a + b}}\right) \geq 9. \] | Solution. Rewrite it into another form\n\n\[ \left( {1 + \frac{a + c}{a + b}}\right) \left( {1 + \frac{a + c}{c + d}}\right) \left( {1 + \frac{b + d}{b + c}}\right) \left( {1 + \frac{b + d}{a + d}}\right) \geq 9. \]\n\nFor all positive real numbers \( x, y \), it’s easy to see that\n\n\[ \left( {1 + \frac{1}{x}}\right)... | Yes |
Problem 31. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{{a}^{2} + {b}^{2}} + \frac{1}{{b}^{2} + {c}^{2}} + \frac{1}{{c}^{2} + {a}^{2}} + \frac{8}{{a}^{2} + {b}^{2} + {c}^{2}} \geq \frac{6}{{ab} + {bc} + {ca}}. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . Denote \( t = \sqrt{{b}^{2} + {c}^{2}} \) and\n\n\[ f\left( {a, b, c}\right) = \frac{1}{{a}^{2} + {b}^{2}} + \frac{1}{{b}^{2} + {c}^{2}} + \frac{1}{{c}^{2} + {a}^{2}} + \frac{8}{{a}^{2} + {b}^{2} + {c}^{2}} - \frac{6}{{ab} + {bc} + {ca}}. \]\n\nWe have\n\n\[ f\left( {... | Yes |
Problem 32. Let \( a, b, c, k \) be positive real numbers and \( k \geq \frac{3}{2} \). Prove that\n\n\[ \frac{{a}^{k}}{a + b} + \frac{{b}^{k}}{b + c} + \frac{{c}^{k}}{c + a} \geq \frac{1}{2}\left( {{a}^{k - 1} + {b}^{k - 1} + {c}^{k - 1}}\right) .\n\]\n\n(Vasile Cirtoaje and Pham Kim Hung) | Solution. I have two solutions to this problem.\n\nFirst Solution. (Cauchy reverse) Rewrite the inequality in the form\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {{a}^{k - 1} - \frac{{a}^{k}}{a + b}}\right) \leq \frac{1}{2}\left( {\mathop{\sum }\limits_{{cyc}}{a}^{k - 1}}\right) \Leftrightarrow \mathop{\sum }\limits_{{c... | Yes |
Problem 33. Let \( a, b, c \) be non-negative real numbers and \( a + b + c = 3 \) . Prove that\n\n\[ a\sqrt{1 + {b}^{3}} + b\sqrt{1 + {c}^{3}} + c\sqrt{1 + {a}^{3}} \leq 5. \] | Solution. By AM-GM inequality, we deduce that\n\n\[ \mathop{\sum }\limits_{{cyc}}a\sqrt{1 + {b}^{3}} = \mathop{\sum }\limits_{{cyc}}a\sqrt{\left( {1 + b}\right) \left( {1 - b + {b}^{2}}\right) } \leq \frac{1}{2}\mathop{\sum }\limits_{{cyc}}a\left( {1 + {b}^{2}}\right) .\n\nIt remains to prove that\n\n\[ a{b}^{2} + b{c}... | Yes |
Problem 35. Suppose that \( a, b, c \) are three non-negative real numbers verifying \( {a}^{2} + {b}^{2} + {c}^{2} = 1 \) . Prove the following inequality\n\n\[ \frac{a}{{a}^{3} + {bc}} + \frac{b}{{b}^{3} + {ac}} + \frac{c}{{c}^{3} + {ab}} \geq 3. \] | Solution. I have two solutions to this problem.\n\nFirst Solution. If all terms in the left hand side are greater than 1 then the inequality is proved immediately. Otherwise, we may assume that\n\n\[ x = \frac{a}{{a}^{3} + {bc}} \leq 1 \Rightarrow a \leq {a}^{3} + {bc}. \]\n\nApplying Cauchy-Schwarz inequality, we obta... | Yes |
Problem 36. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n\frac{b + c}{\sqrt{{a}^{2} + {bc}}} + \frac{c + a}{\sqrt{{b}^{2} + {ca}}} + \frac{a + b}{\sqrt{{c}^{2} + {ab}}} \geq 4.\n\]\n\n(Pham Kim Hung) | Solution. Applying Hölder inequality, we obtain\n\n\[ \n{\left( \mathop{\sum }\limits_{{cyc}}\frac{b + c}{\sqrt{{a}^{2} + {bc}}}\right) }^{2}\left( {\mathop{\sum }\limits_{{cyc}}\left( {b + c}\right) \left( {{a}^{2} + {bc}}\right) }\right) \geq 8{\left( \mathop{\sum }\limits_{{cyc}}a\right) }^{3}.\n\]\n\nTherefore, it'... | Yes |
Problem 37. Let \( a, b, c \) be non-negative real numbers satisfying \( {a}^{2} + {b}^{2} + {c}^{2} + {abc} = 4 \) . Prove that \( 2 + {abc} \geq {ab} + {bc} + {ca} \geq {abc} \) . | Solution. To prove the right hand inequality, just notice that at least one of \( a, b, c \) , say \( a \), is not bigger than 1 . Thereforec we have \( {ab} + {bc} + {ca} \geq {bc} \geq {abc} \) . Equality holds for \( \left( {a, b, c}\right) = \left( {0,0,2}\right) \) up to permutation.\n\nTo prove the right hand ine... | Yes |
Problem 38. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{{a}^{2} + {2bc}}{{b}^{2} + {c}^{2}}} + \sqrt{\frac{{b}^{2} + {2ca}}{{c}^{2} + {a}^{2}}} + \sqrt{\frac{{c}^{2} + {2ab}}{{a}^{2} + {b}^{2}}} \geq 3. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . First, we will prove that\n\n\[ \sqrt{\frac{{a}^{2} + {c}^{2}}{{b}^{2} + {c}^{2}}} + \sqrt{\frac{{b}^{2} + {c}^{2}}{{c}^{2} + {a}^{2}}} \geq \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}}. \]\n\nIndeed, this condition is equivalent to\n\n\[ \frac{{a}^{2} + {c}^{2}}{{b}^{2} +... | Yes |
Problem 39. Let \( a, b, c \) be three distinct positive real numbers. Prove that\n\n\[ \frac{1}{\left| {a}^{2} - {b}^{2}\right| } + \frac{1}{\left| {b}^{2} - {c}^{2}\right| } + \frac{1}{\left| {c}^{2} - {a}^{2}\right| } + \frac{8}{{a}^{2} + {b}^{2} + {c}^{2}} \geq \frac{27}{{\left( a + b + c\right) }^{2}}. \] | Solution. WLOG, we may assume that \( a > b > c \) . Notice that\n\n\[ \frac{1}{{a}^{2} - {b}^{2}} + \frac{1}{{b}^{2} - {c}^{2}} + \frac{1}{{a}^{2} - {c}^{2}} + \frac{8}{{a}^{2} + {b}^{2} + {c}^{2}} - \left( {\frac{1}{{a}^{2} - {b}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{a}^{2}} + \frac{8}{{a}^{2} + {b}^{2}}}\right) \]\n\... | Yes |
Find the best positive real constant \( k \) such that the following inequality holds for all positive real numbers \( a, b \) and \( c \)\n\n\[ \n\\frac{\\left( {a + b}\\right) \\left( {b + c}\\right) \\left( {c + a}\\right) }{abc} + \\frac{k\\left( {{ab} + {bc} + {ca}}\\right) }{{a}^{2} + {b}^{2} + {c}^{2}} \\geq 8 +... | Solution. We clearly have\n\n\[ \n\\frac{\\left( {a + b}\\right) \\left( {b + c}\\right) \\left( {c + a}\\right) }{abc} - 8 = \\frac{c{\\left( a - b\\right) }^{2} + a{\\left( b - c\\right) }^{2} + b{\\left( c - a\\right) }^{2}}{abc},\n\]\n\n\[ \n1 - \\frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} = \\frac{{\\le... | Yes |
Problem 41. Suppose \( a, b, c \) are positive real numbers satisfying the condition \( a + b + c + {abc} = 4 \) . Prove that\n\n\[ \frac{a}{\sqrt{b + c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a + b}} \geq \frac{a + b + c}{\sqrt{2}}. \] | Solution. First we will prove that \( a + b + c \geq {ab} + {bc} + {ca} \) . Indeed, we may suppose that \( c \geq b \geq a \) without loss of generality. We need to prove that\n\n\[ a + b - {ab} \geq \frac{4 - a - b}{{ab} + 1}\left( {a + b - 1}\right) \Leftrightarrow {\left( a + b - 2\right) }^{2} \geq {ab}\left( {a -... | Yes |
Prove that for all non-negative real numbers \( a, b, c \), we have\n\n\[ \sqrt{\frac{2{a}^{2} + {bc}}{{a}^{2} + {2bc}}} + \sqrt{\frac{2{b}^{2} + {ca}}{{b}^{2} + {2ca}}} + \sqrt{\frac{2{c}^{2} + {ab}}{{c}^{2} + {2ab}}} \geq 2\sqrt{2}. \]\n | Since the inequality is homogeneous, we may assume that \( {abc} = 1 \) . The problem becomes\n\n\[ \sqrt{\frac{{2x} + 1}{x + 2}} + \sqrt{\frac{{2y} + 1}{y + 2}} + \sqrt{\frac{{2z} + 1}{z + 2}} \geq 2\sqrt{2} \]\n\nwhere \( x = {a}^{3}, y = {b}^{3}, z = {c}^{2},{xyz} = 1 \) . WLOG, suppose that \( x \geq y \geq z \) . ... | Yes |
Problem 43. Let \( x, y, z \) be non-negative real numbers with sum 1. Prove that\n\n\[ \sqrt{x + \frac{{\left( y - z\right) }^{2}}{12}} + \sqrt{y + \frac{{\left( z - x\right) }^{2}}{12}} + \sqrt{z + \frac{{\left( x - y\right) }^{2}}{12}} \leq \sqrt{3}. \]\n\n\[ \text{(Phan Thanh Nam, VMEO 2004)} \] | Solution. Suppose \( z = \min \{ x, y, z\} \) . First we will prove that if \( u = y - z, v = x - z \) and \( k = \frac{1}{12} \) then\n\n\[ \sqrt{x + k{u}^{2}} + \sqrt{y + k{v}^{2}} \leq \sqrt{2\left( {x + y}\right) + k{\left( u + v\right) }^{2}}. \]\n\nIndeed, this one is equivalent to\n\n\[ 2\sqrt{\left( {x + k{u}^{... | Yes |
Let \( x, y, z \) be three non-negative real numbers satisfying the condition \( {xy} + {yz} + {zx} = 1 \) . Prove that\n\n\[ \frac{1}{\sqrt{x + y}} + \frac{1}{\sqrt{y + z}} + \frac{1}{\sqrt{z + x}} \geq 2 + \frac{1}{\sqrt{2}}. \] | Solution. WLOG, we may assume that \( x = \max \left( {x, y, z}\right) \) . Denote \( a = y + z > 0 \) , then obviously, \( {ax} = 1 - {yz} \leq 1 \) . Consider the function\n\n\[ f\left( x\right) = \frac{1}{\sqrt{x + y}} + \frac{1}{\sqrt{y + z}} + \frac{1}{\sqrt{z + x}} \]\n\n\[ = \frac{1}{\sqrt{y + z}} + \sqrt{\frac{... | Yes |
Problem 46. Let \( a, b, c \) be non-negative real numbers and \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \frac{1}{3 - {ab}} + \frac{1}{3 - {bc}} + \frac{1}{3 - {ca}} + \frac{1}{3 - {a}^{2}} + \frac{1}{3 - {b}^{2}} + \frac{1}{3 - {c}^{2}} \geq 3. \] | Solution. Rewrite the inequality in the form\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{3}{3 - {ab}} - 1}\right) + \mathop{\sum }\limits_{{cyc}}\left( {\frac{3}{3 - {c}^{2}} - 1}\right) \geq 3 \]\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{ab}{3 - {ab}} + \mathop{\sum }\limits_{{cyc}}\frac{{c}^{2}}{3... | Yes |
Problem 47. Let \( a, b, c \) be three positive real numbers. Prove that\n\n\[ \frac{1}{a\sqrt{a + b}} + \frac{1}{b\sqrt{b + c}} + \frac{1}{c\sqrt{c + a}} \geq \frac{3}{\sqrt{2abc}} \]\n\n(Phan Thanh Nam, VMEO 2005) | Solution. Let \( x = \frac{\sqrt{2bc}}{\sqrt{a\left( {a + b}\right) }}, y = \frac{\sqrt{2ca}}{\sqrt{b\left( {b + c}\right) }}, z = \frac{\sqrt{2ab}}{\sqrt{c\left( {c + a}\right) }} \) . We need to prove that \( x + y + z \geq 3 \) . However, it suffices to prove the following stronger result as follow\n\n\[ 3 \leq {xy}... | Yes |
Problem 48. Prove that \( {ax} + {by} + {cz} \geq 0 \) if \( a, b, c, x, y, z \) are real numbers such that\n\n\[ \n\left( {a + b + c}\right) \left( {x + y + z}\right) = 3\;;\;\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) = 4.\n\]\n\n(Mathlinks Contest) | Solution. Let \( \alpha = \sqrt[4]{\frac{{a}^{2} + {b}^{2} + {c}^{2}}{{x}^{2} + {y}^{2} + {z}^{2}}} \) and \( {a}_{1} = \frac{a}{\alpha },{b}_{1} = \frac{b}{\alpha },{c}_{1} = \frac{c}{\alpha },{x}_{1} = {x\alpha },{y}_{1} = \n\n\( {y\alpha },{z}_{1} = {z\alpha } \) . We infer that\n\n\[ \n{a}_{1}^{2} + {b}_{1}^{2} + {... | Yes |
Problem 49. Let \( a, b, c, d \) be non-negative real numbers with sum 4. Prove that\n\n\[ \sqrt{\frac{a + 1}{{ab} + 1}} + \sqrt{\frac{b + 1}{{bc} + 1}} + \sqrt{\frac{c + 1}{{cd} + 1}} + \sqrt{\frac{d + 1}{{da} + 1}} \geq 4. \] | Solution. According to AM-GM inequality, we get\n\n\[ \text{ LHS } \geq 4\sqrt[8]{\frac{\left( {a + 1}\right) \left( {b + 1}\right) \left( {c + 1}\right) \left( {b + 1}\right) }{\left( {{ab} + 1}\right) \left( {{bc} + 1}\right) \left( {{cd} + 1}\right) \left( {{da} + 1}\right) }}, \]\n\nand it remains to prove that\n\n... | Yes |
Prove that for all non-negative real numbers \( a, b, c \) then\n\n\[ \frac{a}{\sqrt{a + b}} + \frac{b}{\sqrt{b + c}} + \frac{c}{\sqrt{c + a}} \geq \frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{2}}. \] | Solution. Let \( x = \sqrt{a}, y = \sqrt{b}, z = \sqrt{c} \) . The inequality becomes\n\n\[ \frac{{x}^{2}}{\sqrt{{x}^{2} + {y}^{2}}} + \frac{{y}^{2}}{\sqrt{{y}^{2} + {z}^{2}}} + \frac{{z}^{2}}{\sqrt{{z}^{2} + {x}^{2}}} \geq \frac{x + y + z}{\sqrt{2}} \]\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{2{x}^{4}}... | Yes |
Let \( a, b, c \) be real numbers. Prove that\n\n\[ \sqrt[3]{2{a}^{2} - {bc}} + \sqrt[3]{2{b}^{2} - {ca}} + \sqrt[3]{2{c}^{2} - {ab}} \geq 0. \] | Solution. First, we notice that the inequality only needs to be considered for \( a, b, c \) non-negative. Now, consider the identity\n\n\[ {x}^{3} + {y}^{3} + {z}^{3} - {3xyz} = \left( {x + y + z}\right) \left( {{x}^{2} + {y}^{2} + {z}^{2} - {xy} - {yz} - {zx}}\right) \]\n\ntherefore, \( \left( {x + y + z}\right) \lef... | Yes |
Prove that the following inequality holds for all real numbers \( a, b, c \)\n\n\[ \n{\\left( {a}^{2} + {b}^{2} + {c}^{2}\\right) }^{2} \\geq 3\\left( {{a}^{3}b + {b}^{3}c + {c}^{3}a}\\right) .\n\] | Solution. We give four solutions to this problem.\n\nFirst Solution. Notice that\n\n\[ \n4\\left( {{a}^{2} + {b}^{2} + {c}^{2} - {ab} - {bc} - {ca}}\\right) \\left( {{\\left( {a}^{2} + {b}^{2} + {c}^{2}\\right) }^{2} - 3\\left( {{a}^{3}b + {b}^{3}c + {c}^{3}a}\\right) }\\right)\n\]\n\n\[ \n= {\\left( \\left( {a}^{3} + ... | Yes |
Problem 53. Let \( a, b, c \) be three real numbers satisfying the condition \( {a}^{2} + {b}^{2} + {c}^{2} = 9 \) . Prove that\n\n\[ 3\min \left( {a, b, c}\right) \leq 1 + {abc}. \] | Solution. WLOG, we may assume that \( c \geq b \geq a \) . Consider the following cases\n\n(i). \( a \leq 0 \) : Let \( d = - a \) and \( e = \left| b\right| \) . We will prove that\n\n\[ - {3d} \leq 1 - {dce} \Leftrightarrow d\left( {{ce} - 3}\right) \leq 1. \]\n\nIf \( {ce} \leq 3 \), the conclusion follows immediate... | Yes |
Problem 54. Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \left( {1 + {a}^{4}}\right) \left( {1 + {b}^{4}}\right) \left( {1 + {c}^{4}}\right) \left( {1 + {d}^{4}}\right) \geq \left( {1 + {a}^{3}}\right) \left( {1 + {b}^{3}}\right) \left( {1 + {c}^{3}}\right) \lef... | Solution. Notice that for all \( x \geq 0,\left( {1 + {x}^{4}}\right) \left( {1 + x}\right) \geq \left( {1 + {x}^{3}}\right) \left( {1 + {x}^{2}}\right) \), therefore\n\n\[ \mathop{\prod }\limits_{{cyc}}\left( {1 + {a}^{4}}\right) \mathop{\prod }\limits_{{cyc}}\left( {1 + a}\right) \geq \mathop{\prod }\limits_{{cyc}}\l... | Yes |
Problem 56. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3\sqrt{\frac{{a}^{2} + {b}^{2} + {c}^{2}}{{ab} + {bc} + {ca}}}.\n\] | Solution. Notice that if \( a \geq b \geq c \) then\n\n\[ \n\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}\right) - \left( {\frac{b}{a} + \frac{c}{b} + \frac{a}{c}}\right) = \frac{\left( {a - b}\right) \left( {a - c}\right) \left( {c - b}\right) }{abc} \leq 0 \n\]\n\nso it’s enough to consider the case \( a \geq b \g... | Yes |
Suppose that \( n \) positive real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) satisfy the condition\n\n\[ \frac{1}{1 + {x}_{1}} + \frac{1}{1 + {x}_{2}} + \ldots + \frac{1}{1 + {x}_{n}} = \frac{n}{2}. \]\n\nProve that\n\n\[ \mathop{\sum }\limits_{{i, j = 1}}^{n}\frac{1}{{x}_{i} + {x}_{j}} \geq \frac{{n}^{2}}{2} \] | For each \( i \in \{ 1,2,\ldots, n\} \), we denote \( {a}_{i} = \frac{1 - {x}_{i}}{1 + {x}_{i}} \) . Therefore, \( {a}_{1} + {a}_{2} + \ldots + \) \( {a}_{n} = 0 \) and \( {a}_{i} \in \left\lbrack {-1,1}\right\rbrack \) . Consider the expression\n\n\[ S = \mathop{\sum }\limits_{{i, j = 1}}^{n}\frac{1}{{x}_{i} + {x}_{j}... | Yes |
Problem 59. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{ab}{a + {4b} + {4c}} + \frac{bc}{b + {4c} + {4a}} + \frac{ca}{c + {4a} + {4b}} \leq \frac{a + b + c}{9}. \]\n\n(Pham Kim Hung) | Solution. WLOG, we may assume that \( a + b + c = 3 \) . The inequality becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{3ab}{a + 4\left( {3 - a}\right) } \leq 1 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{ab}{4 - a} \leq 1 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{b}{4 - a} \leq 1 \]\n\n\[ \Leftrightar... | Yes |
Problem 60. Suppose that \( n \) is an integer greater than 2. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers such that \( {a}_{1}{a}_{2}\ldots {a}_{n} = 1 \) . Prove the following inequality\n\n\[ \frac{{a}_{1} + 3}{{\left( {a}_{1} + 1\right) }^{2}} + \frac{{a}_{2} + 3}{{\left( {a}_{2} + 1\right) }... | Solution. Notice first that it is sufficient to prove the inequality in the case \( n = 3 \) . For a bigger value of \( n\left( {n \geq 4}\right) \), we only need to choose from the set \( \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) the three smallest numbers, say \( {a}_{1},{a}_{2},{a}_{3} \) . Since \( {a}_{... | Yes |
Problem 61. Let \( a, b, c \) be non-negative real numbers with sum 2. Prove that\n\n\[ \sqrt{a + b - {2ab}} + \sqrt{b + c - {2bc}} + \sqrt{c + a - {2ca}} \geq 2. \] | Solution. WLOG, we may assume that \( a \geq b \geq c \) . Let \( x = a + b - {2ab}, y = b + c - {2bc} \) and \( z = c + a - {2ca} \) . The inequality is equivalent to (after squaring)\n\n\[ 2\mathop{\sum }\limits_{{cyc}}\sqrt{xy} \geq 2\mathop{\sum }\limits_{{cyc}}{ab} \]\n\nNotice that \( {2x} = c\left( {a + b}\right... | Yes |
Problem 62. Let \( a, b, c \) be non-negative real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \frac{a}{b + 2} + \frac{b}{c + 2} + \frac{c}{a + 2} \leq 1. \] | Solution. After expanding, the inequality gets a simpler form\n\n\[ a{b}^{2} + b{c}^{2} + c{a}^{2} \leq 2 + {abc}. \]\n\nWLOG, suppose that \( b \) is the second greatest number in the set \( \{ a, b, c\} \), then\n\n\[ a\left( {b - a}\right) \left( {b - c}\right) \leq 0 \Leftrightarrow {a}^{2}b + {abc} \geq a{b}^{2} +... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a\left( {b + c}\right) }{{a}^{2} + {bc}} + \frac{b\left( {c + a}\right) }{{b}^{2} + {ca}} + \frac{c\left( {a + b}\right) }{{c}^{2} + {ab}} \geq 2. \] | The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}a\left( {b + c}\right) \left( {{b}^{2} + {ca}}\right) \left( {{c}^{2} + {ab}}\right) \geq 2\left( {{a}^{2} + {bc}}\right) \left( {{b}^{2} + {ca}}\right) \left( {{c}^{2} + {ab}}\right) \]\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{{cyc}}{a}^{4}\left( {... | Yes |
Problem 64. Suppose that \( a, b, c \) are the side lengths of a triangle. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} + \frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} \leq \frac{5}{2}. \]\n | Solution. With the following identities\n\n\[ - 3 + \mathop{\sum }\limits_{{cyc}}\frac{2a}{b + c} = \mathop{\sum }\limits_{{cyc}}\frac{{\left( a - b\right) }^{2}}{\left( {a + c}\right) \left( {b + c}\right) } \]\n\n\[ 2 - \frac{2\left( {{ab} + {bc} + {ca}}\right) }{{a}^{2} + {b}^{2} + {c}^{2}} = \frac{{\left( a - b\rig... | Yes |
Problem 65. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{\sqrt{{a}^{2} + {bc}}} + \frac{1}{\sqrt{{b}^{2} + {ca}}} + \frac{1}{\sqrt{{c}^{2} + {ab}}} \geq \frac{6}{a + b + c}. \]\n(Pham Kim Hung) | Solution. First solution. Taking into account problem 15, we have\n\n\[ \frac{1}{\sqrt{{a}^{2} + {bc}}} + \frac{1}{\sqrt{{b}^{2} + {ca}}} + \frac{1}{\sqrt{{c}^{2} + {ab}}} \geq \frac{9}{\sqrt{{a}^{2} + {bc}} + \sqrt{{b}^{2} + {ca}} + \sqrt{{c}^{2} + {ab}}} \geq \frac{6}{a + b + c}. \] | Yes |
Problem 66. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{3}}{2{a}^{2} - {ab} + 2{b}^{2}} + \frac{{b}^{3}}{2{b}^{2} - {bc} + 2{c}^{2}} + \frac{{c}^{3}}{2{c}^{2} - {ca} + 2{a}^{2}} \geq \frac{a + b + c}{3}. \]\n\n(Nguyen Viet Anh) | Solution. Rewrite the inequality form as follows\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{2{a}^{2} - {ab} + 2{b}^{2}} - \frac{1}{3}\mathop{\sum }\limits_{{cyc}}a = \mathop{\sum }\limits_{{cyc}}\frac{a\left( {{a}^{2} + {ab} - 2{b}^{2}}\right) }{3\left( {2{a}^{2} - {ab} + 2{b}^{2}}\right) }\n\n\[ = \mathop{\sum ... | Yes |
Prove that for all positive real numbers \( a, b, c \)\n\n\[ \frac{{a}^{2}}{b} + \frac{{b}^{2}}{c} + \frac{{c}^{2}}{a} \geq 3\sqrt[4]{\frac{{a}^{4} + {b}^{4} + {c}^{4}}{3}}. \] | Solution. Applying Hölder inequality, we obtain\n\n\[ \left( {\frac{{a}^{2}}{b} + \frac{{b}^{2}}{c} + \frac{{c}^{2}}{a}}\right) \left( {\frac{{a}^{2}}{b} + \frac{{b}^{2}}{c} + \frac{{c}^{2}}{a}}\right) \left( {{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}}\right) \geq {\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{3}... | Yes |
Problem 69. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{\left( a + b\right) }^{2}}{{c}^{2} + {ab}} + \frac{{\left( b + c\right) }^{2}}{{a}^{2} + {bc}} + \frac{{\left( c + a\right) }^{2}}{{b}^{2} + {ac}} \geq 6. \] | Solution. We have \( {\left( a + b\right) }^{2} - 2\left( {{c}^{2} + {ab}}\right) = \left( {{a}^{2} - {c}^{2}}\right) + \left( {{b}^{2} - {c}^{2}}\right) \), therefore\n\n\[ \frac{{\left( a + b\right) }^{2}}{{c}^{2} + {ab}} + \frac{{\left( b + c\right) }^{2}}{{a}^{2} + {bc}} + \frac{{\left( c + a\right) }^{2}}{{b}^{2} ... | Yes |
Find the maximum value of \( k = k\\left( n\\right) \) for which the following inequality is true for all real numbers \( {x}_{1},{x}_{2},\\ldots ,{x}_{n} \)\n\n\\[ \n{x}_{1}^{2} + {\\left( {x}_{1} + {x}_{2}\\right) }^{2} + \\ldots + {\\left( {x}_{1} + {x}_{2} + \\ldots + {x}_{n}\\right) }^{2} \\geq k{\\left( {x}_{1} +... | Let \( {a}_{1},{a}_{2},\\ldots ,{a}_{n} \) be positive real numbers, then\n\n\\[ \n{a}_{1}{y}_{1}^{2} + \\frac{1}{{a}_{1}} \\cdot {y}_{2}^{2} + 2{y}_{1}{y}_{2} \\geq 0\n\\]\n\n\\[ \n{a}_{2}{y}_{2}^{2} + \\frac{1}{{a}_{2}} \\cdot {y}_{3}^{2} + 2{y}_{2}{y}_{3} \\geq 0\n\\]\n\n\\( \\ldots \\;\\ldots \\)\n\n\\[ \n{a}_{n - ... | Yes |
Problem 72. Let \( x, y, z \) be non-negative real numbers with sum 1. Prove that\n\n\[ \sqrt{x + {y}^{2}} + \sqrt{y + {z}^{2}} + \sqrt{z + {x}^{2}} \geq 2. \] | (Phan Thanh Nam) Solution. Notice that if \( a, b, c, d \) are non-negative real numbers such that \( a + b = c + d \) and \( \left| {a - b}\right| \leq \left| {c - d}\right| \), then we have\n\n\[ \sqrt{a} + \sqrt{b} \geq \sqrt{c} + \sqrt{d}\left( \star \right) \]\n\nIndeed, since \( {\left( a + b\right) }^{2} - {\lef... | Yes |
Problem 73. Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \frac{1}{2 + {a}^{2}{b}^{2}} + \frac{1}{2 + {b}^{2}{c}^{2}} + \frac{1}{2 + {c}^{2}{a}^{2}} \geq 1. \] | Solution. According to AM-GM inequality, we have\n\n\[ \frac{1}{2 + {a}^{2}{b}^{2}} = \frac{1}{2} - \frac{{a}^{2}{b}^{2}}{2\left( {2 + {a}^{2}{b}^{2}}\right) } \geq \frac{1}{2} - \frac{{a}^{2}{b}^{2}}{6\sqrt[3]{{a}^{2}{b}^{2}}} = \frac{1}{2} - \frac{{a}^{4/3}{b}^{4/3}}{6}. \]\n\nWe deduce that\n\n\[ \mathop{\sum }\limi... | Yes |
Consider the positive real constants \( m, n \), such that \( 3{n}^{2} > {m}^{2} \). For real numbers \( a, b, c \) such that \( a + b + c = m,{a}^{2} + {b}^{2} + {c}^{2} = {n}^{2} \), find the maximum and minimum of\n\n\[ P = {a}^{2}b + {b}^{2}c + {c}^{2}a. \] | Solution. Let \( a = x + \frac{m}{3}, b = y + \frac{m}{3}, c = z + \frac{m}{3} \). From the given conditions, we get that \( x + y + z = 0 \) and \( {x}^{2} + {y}^{2} + {z}^{2} = \frac{3{n}^{2} - {m}^{2}}{3} \). The expression \( P \) becomes\n\n\[ P = {x}^{2}y + {y}^{2}z + {z}^{2}x + \frac{{m}^{3}}{9}. \]\n\nNotice th... | Yes |
Problem 75. Suppose that \( a, b, c \) are three positive real numbers verifying\n\n\[ \n\\left( {a + b + c}\\right) \\left( {\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}}\\right) = {13}.\n\]\n\nFind the minimum and maximum values of the expression\n\n\[ \nP = \\frac{{a}^{3} + {b}^{3} + {c}^{3}}{abc}.\n\]\n\n(Pham Kim Hu... | Solution. Denote\n\n\[ \nx = \\mathop{\\sum }\\limits_{{cyc}}\\frac{a}{b};y = \\mathop{\\sum }\\limits_{{cyc}}\\frac{b}{a};m = \\mathop{\\sum }\\limits_{{cyc}}\\frac{{a}^{2}}{bc};n = \\mathop{\\sum }\\limits_{{cyc}}\\frac{bc}{{a}^{2}};\n\]\n\nWe have \( x + y = {10} \) and\n\n\[ \n{x}^{3} = 3\\left( {m + n}\\right) + 6... | Yes |
Prove that for all positive real numbers \( a, b, c, d, e \) ,\n\n\[ \n\frac{a + b}{2} \cdot \frac{b + c}{2} \cdot \frac{c + d}{2} \cdot \frac{d + e}{2} \cdot \frac{e + a}{2} \leq \frac{a + b + c}{3} \cdot \frac{b + c + d}{3} \cdot \frac{c + d + e}{3} \cdot \frac{d + e + a}{3} \cdot \frac{e + a + b}{3}.\n\] | Solution. We will first prove that for all \( a, b > 0 \) and \( a + b \leq 1 \)\n\n\[ \n\left( {\frac{1}{a} - 1}\right) \left( {\frac{1}{b} - 1}\right) \geq {\left( \frac{2}{a + b} - 1\right) }^{2}\left( \star \right)\n\]\n\nIndeed, this result can be rewritten in the following form\n\n\[ \n\frac{1}{ab} - \frac{1}{a} ... | Yes |
Problem 77. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{4}}{{a}^{3} + {b}^{3}} + \frac{{b}^{4}}{{b}^{3} + {c}^{3}} + \frac{{c}^{4}}{{c}^{3} + {a}^{3}} \geq \frac{a + b + c}{2}. \]\n | Solution. Notice that\n\n\[ \frac{2{a}^{4}}{{a}^{3} + {b}^{3}} - a - \frac{3\left( {a - b}\right) }{2} = \left( {a - b}\right) \left( {\frac{a\left( {{a}^{2} + {ab} + {b}^{2}}\right) }{{a}^{3} + {b}^{3}} - \frac{3}{2}}\right) = \frac{2{b}^{2} + {ab} - {b}^{2}}{3\left( {{a}^{3} + {b}^{3}}\right) }{\left( a - b\right) }^... | Yes |
Problem 78. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \sqrt{\frac{{a}^{3}}{{a}^{2} + {ab} + {b}^{2}}} + \sqrt{\frac{{b}^{3}}{{b}^{2} + {bc} + {c}^{2}}} + \sqrt{\frac{{c}^{3}}{{c}^{2} + {ca} + {a}^{2}}} \geq \frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{3}}. \] | Solution. Let \( {x}^{2} = a,{y}^{2} = b \) and \( {z}^{2} = c \) . The inequality becomes\n\n\[ \frac{{x}^{3}}{\sqrt{{x}^{4} + {x}^{2}{y}^{2} + {y}^{4}}} + \frac{{y}^{3}}{\sqrt{{y}^{4} + {y}^{2}{z}^{2} + {z}^{4}}} + \frac{{z}^{3}}{\sqrt{{z}^{4} + {z}^{2}{x}^{2} + {x}^{4}}} \geq \frac{x + y + z}{\sqrt{3}}. \]\n\nSquari... | Yes |
Problem 81. Let \( x, y, z \) be positive real numbers satisfying \( {2xyz} = 3{x}^{2} + 4{y}^{2} + 5{z}^{2} \). Find the minimum of the expression \( P = {3x} + {2y} + z \) . | Solution. Let \( a = {3x}, b = {2y}, z = c \) We then obtain \[ a + b + c = {3x} + {2y} + z,{a}^{2} + 3{b}^{2} + {15}{c}^{2} = {abc}. \] According to the weighted AM-GM inequality, we have that \[ a + b + c \geq {\left( 2a\right) }^{1/2}{\left( 3b\right) }^{1/3}{\left( 6c\right) }^{1/6}, \] \[ {a}^{2} + 3{b}^{2} + {15}... | Yes |
Problem 82. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{\sqrt{{a}^{2} + {bc}}} + \frac{1}{\sqrt{{b}^{2} + {ca}}} + \frac{1}{\sqrt{{c}^{2} + {ab}}} \geq \frac{2\sqrt{2}}{\sqrt{{ab} + {bc} + {ca}}}. \]\n\n(Pham Kim Hung) | Solution. First we may assume that \( a \geq b \geq c \) . Notice that\n\n\[ \frac{1}{\sqrt{{b}^{2} + {ca}}} + \frac{1}{\sqrt{{c}^{2} + {ab}}} \geq \frac{2\sqrt{2}}{\sqrt{{b}^{2} + {c}^{2} + {ab} + {ac}}} \]\n\nso it suffices to prove that\n\n\[ \frac{1}{\sqrt{{a}^{2} + {bc}}} + \frac{2\sqrt{2}}{\sqrt{{b}^{2} + {c}^{2}... | Yes |
Problem 83. Let \( a, b, c, d \) be positive real numbers with sum 4. Prove that\n\n\[ \frac{1}{5 - {abc}} + \frac{1}{5 - {bcd}} + \frac{1}{5 - {cda}} + \frac{1}{5 - {abc}} \leq 1. \] | Solution. Let \( x = {abc}, y = {bcd}, z = {cda}, t = {dab} \) . We need to prove that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{5 - x} \leq 1 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{1 - x}{5 - x} \geq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{\left( {1 - x}\right) \left( {x + 2}\right) }{\left(... | Yes |
Problem 84. Let \( a, b, c \) be three arbitrary real numbers. Prove that\n\n\[ \n\frac{1}{{\left( 2a - b\right) }^{2}} + \frac{1}{{\left( 2b - c\right) }^{2}} + \frac{1}{{\left( 2c - a\right) }^{2}} \geq \frac{11}{7\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) }.\n\]\n\n(Pham Kim Hung) | Solution. Denote \( x = {2a} - b, y = {2b} - c, z = {2c} - a \) . We get\n\n\[ a = \frac{{4x} + {2y} + z}{7}, b = \frac{{4y} + {2z} + x}{7}, c = \frac{{4z} + {2x} + y}{7},\]\n\n\[ \Rightarrow {a}^{2} + {b}^{2} + {c}^{2} = \frac{2{\left( x + y + z\right) }^{2} + {x}^{2} + {y}^{2} + {z}^{2}}{7}. \]\n\nIt remains to prove... | Yes |
Problem 85. Let \( a, b, c \) be positive real numbers. Consider the following inequality\n\n\[ \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} \geq 3\sqrt[k]{\frac{{a}^{k} + {b}^{k} + {c}^{k}}{3}}\left( \star \right) \]\n\n(a). Prove that \( \left( \star \right) \) is true for \( k = 2 \) . | Solution. (a). For \( k = 2 \), we can assume that \( \mathop{\sum }\limits_{{cyc}}{a}^{2} = 3 \) without loss of generality. The inequality \( \mathop{\sum }\limits_{{cyc}}\frac{ab}{c} \geq 3 \) is equivalent to\n\n\[ {\left( \mathop{\sum }\limits_{{cyc}}\frac{ab}{c}\right) }^{2} \geq 9 \Leftrightarrow \mathop{\sum }\... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{\sqrt{4{a}^{2} + {bc}}} + \frac{1}{\sqrt{4{b}^{2} + {ca}}} + \frac{1}{\sqrt{4{c}^{2} + {ab}}} \geq \frac{4}{a + b + c}. \] | Solution. First Solution. We denote\n\n\[ S = \mathop{\sum }\limits_{{cyc}}\frac{1}{\sqrt{4{a}^{2} + {bc}}};\;P = \mathop{\sum }\limits_{{cyc}}{\left( b + c\right) }^{3}\left( {4{a}^{2} + {bc}}\right) . \]\n\nAccording to Hölder inequality, we deduce that\n\n\[ S \cdot S \cdot P \geq {\left( a + b + c\right) }^{3}. \]\... | Yes |
Problem 87. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{ab}{4{a}^{2} + {b}^{2} + 4{c}^{2}}} + \sqrt{\frac{bc}{4{b}^{2} + {c}^{2} + 4{a}^{2}}} + \sqrt{\frac{ca}{4{c}^{2} + {a}^{2} + 4{b}^{2}}} \leq 1. \]\n\n(Pham Kim Hung) | Solution. WLOG, suppose that \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . By the weighted Jensen inequality, we deduce that\n\n\[ \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{ab}{4{a}^{2} + {b}^{2} + 4{c}^{2}}} = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2} + 4{b}^{2} + 4{c}^{2}}{27} \cdot \sqrt{\frac{{27}^{2} \cdot {ab}}{\left( ... | Yes |
Problem 88. Suppose that \( n \) is a positive integer and \( \left( {{x}_{1},{x}_{2},..,{x}_{n}}\right) ;\left( {{y}_{1},{y}_{2},...,{y}_{n}}\right) \) are two positive real sequences. Let \( \left( {{z}_{2},{z}_{3},\ldots ,{z}_{2n}}\right) \) be a positive sequence satisfying\n\n\[ \n{z}_{i + j}^{2} \geq {x}_{i}{y}_{... | Solution. Let \( X = \max \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \) and \( Y = \max \left\{ {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right\} \) . WLOG, we can assume that \( X = Y = 1 \) (otherwise, replace \( {x}_{i} \) with \( {x}_{i}/X,{y}_{i} \) with \( {y}_{i}/Y \) and \( {z}_{i} \) with \( \left. {{z}_{i}/\sq... | Yes |
Problem 89. (a).Let \( a, b, c \) be three real numbers. Prove that\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} + a{b}^{3} + b{c}^{3} + c{a}^{3} \geq 2\left( {{a}^{3}b + {b}^{3}c + {c}^{3}a}\right) .\n\] | Solution. For all real numbers \( a, b, c \), we have that\n\n\[ \n{\\left( {a}^{2} - kab + kac - {c}^{2}\\right) }^{2} + {\\left( {b}^{2} - kbc + kba - {a}^{2}\\right) }^{2} + {\\left( {c}^{2} - kca + kcb - {b}^{2}\\right) }^{2} \geq 0.\n\]\n\nAfter expanding, this inequality becomes\n\n\[ \n\\mathop{\\sum }\\limits_{... | Yes |
Problem 91. Let \( a, b, c \) be positive real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \n{a}^{3}{b}^{2} + {b}^{3}{c}^{2} + {c}^{3}{a}^{2} \leq 3 \n\] | Solution. By Cauchy-Schwarz inequality, we have\n\n\[ \n{\left( {a}^{3}{b}^{2} + {b}^{3}{c}^{2} + {c}^{3}{a}^{2}\right) }^{2} \leq \left( {{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}}\right) \left( {{a}^{4}{b}^{2} + {b}^{4}{c}^{2} + {c}^{4}{a}^{2}}\right) .\n\]\n\nIt remains to prove that if \( x + y + z = 3 \) th... | Yes |
Problem 92. Let \( a, b, c \) be arbitrary positive real numbers. Prove that\n\n\[{\left( a + \frac{{b}^{2}}{c}\right) }^{2} + {\left( b + \frac{{c}^{2}}{a}\right) }^{2} + {\left( c + \frac{{a}^{2}}{b}\right) }^{2} \geq \frac{{12}\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) }{a + b + c}.\] | Solution. The inequality is equivalent to\n\n\[{a}^{2} + {b}^{2} + {c}^{2} + \frac{{2a}{b}^{2}}{c} + \frac{{2b}{c}^{2}}{a} + \frac{{2c}{a}^{2}}{b} + \frac{{b}^{4}}{{c}^{2}} + \frac{{c}^{4}}{{a}^{2}} + \frac{{a}^{4}}{{b}^{2}} \geq \frac{{12}\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) }{a + b + c}.\]\n\nUsing the followi... | Yes |
Problem 93. Suppose that \( n \) is an integer greater than 2 and \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are \( n \) real numbers. Prove that for any non-empty subset \( S \) of \( \{ 1,2,\ldots, n\} \), we have\n\n\[ \n{\left( \mathop{\sum }\limits_{{i \in S}}{a}_{i}\right) }^{2} \leq \mathop{\sum }\limits_{{1 \leq i \... | Solution. We will first prove the following lemma\n\nLemma. For all real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{{2k} + 1} \)\n\n\[ \n{\left( \mathop{\sum }\limits_{{0 \leq i \leq k}}{a}_{{2i} + 1}\right) }^{2} \leq \mathop{\sum }\limits_{{1 \leq i \leq j \leq {2k} + 1}}{\left( {a}_{i} + \ldots + {a}_{j}\right) }^{2}\l... | No |
Problem 98. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{{\left( 2a + b\right) }^{2}} + \frac{1}{{\left( 2b + c\right) }^{2}} + \frac{1}{{\left( 2c + a\right) }^{2}} \geq \frac{1}{{ab} + {bc} + {ca}}. \]\n\n(Pham Kim Hung) | Solution. After expanding, the inequality becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {4{a}^{5}b + 4{a}^{5}c - {12}{a}^{4}{b}^{2} + {12}{a}^{4}{c}^{2} + 5{a}^{3}{b}^{3} + + 8{a}^{4}{bc} - {19}{a}^{3}{b}^{2}c + 5{a}^{3}{c}^{2}b - 7{a}^{2}{b}^{2}{c}^{2}}\right) \geq 0 \]\n\nor\n\n\[ 6\mathop{\sum }\limits_{{cyc}}{a... | Yes |
Problem 99. Suppose that \( {x}_{1} \geq {x}_{2} \geq \ldots \geq {x}_{{2n} - 1} \geq {x}_{2n} \geq 0 \) are real numbers and \( {x}_{1} + {x}_{2} + \ldots + {x}_{2n} = {2n} - 1 \) . Find the maximum of the following expression\n\n\[ P = \left( {{x}_{1}^{2} + {x}_{2}^{2}}\right) \left( {{x}_{3}^{2} + {x}_{4}^{2}}\right... | Solution. Although it's very hard to solve this problem directly, we find out unexpectedly that proving the general problem is simpler. In fact, the proposed problem is a direct corollary of the following general result\n\n\( \# \;{Suppose}\;{that}\;\epsilon \leq \frac{k}{2n}\;{is}\;a\;{positive}\;{constant}\;{and}\;{x... | No |
Lemma 1. Let \( x \geq y \geq z \geq t \geq 0 \) and \( y + z = {2\alpha } \) then\n\n\[ \left( {{x}^{2} + {y}^{2}}\right) \left( {{z}^{2} + {t}^{2}}\right) \leq \left( {{x}^{2} + {\alpha }^{2}}\right) \left( {{\alpha }^{2} + {t}^{2}}\right) . \] | Proof. Let \( y = \alpha + \beta \) and \( z = \alpha - \beta ,\beta \geq 0 \), then \( x \geq \alpha + \beta \geq \alpha - \beta \geq t \) . Denote\n\n\[ f\left( \beta \right) = \left\lbrack {{x}^{2} + {\left( \alpha + \beta \right) }^{2}}\right\rbrack \left\lbrack {{\left( \alpha - \beta \right) }^{2} + {t}^{2}}\righ... | Yes |
Lemma 2. Let \( x \geq y \geq z \geq 0 \) and \( \left( {{2n} - 1}\right) x + {2y} = \left( {{2n} + 1}\right) \gamma \left( {n \in \mathbb{N}, n \geq 2}\right) \) then\n\n\[ \n{x}^{{2n} - 2}\left( {{x}^{2} + {y}^{2}}\right) \left( {{y}^{2} + {z}^{2}}\right) \leq 2{\gamma }^{2n}\left( {{\gamma }^{2} + {z}^{2}}\right) .\... | Proof. There exists a real number \( \beta \geq 0 \) for which \( x = \gamma + {2\beta } \) and \( y = \gamma - \left( {{2n} - 1}\right) \beta \) .\n\nThus, we must have \( \gamma - \left( {{2n} - 1}\right) \beta \geq z \) . Denote\n\n\[ \ng\left( \beta \right) = {\left( \gamma + 2\beta \right) }^{2n}{\left( \gamma - \... | Yes |
Theorem 1.1 If \( a, b, c \) are arbitrary real numbers such that \( a + b + c = 1 \), then setting \( {ab} + {bc} + {ca} = \frac{1 - {q}^{2}}{3}\left( {q \geq 0}\right) \), the following inequality holds | \[ \frac{{\left( 1 + q\right) }^{2}\left( {1 - {2q}}\right) }{27} \leq {abc} \leq \frac{{\left( 1 - q\right) }^{2}\left( {1 + {2q}}\right) }{27}. \] | Yes |
Theorem 1 (Generalized Schur Inequality). Let \( a, b, c, x, y, z \) be six non-negative real numbers such that the sequences \( \left( {a, b, c}\right) \) and \( \left( {x, y, z}\right) \) are monotone, then\n\n\[ x\left( {a - b}\right) \left( {a - c}\right) + y\left( {b - a}\right) \left( {b - c}\right) + z\left( {c ... | Proof. WLOG, assume that \( a \geq b \geq c \) . Consider the following cases\n\n(i). \( x \geq y \geq z \) . Then, we have \( \left( {c - a}\right) \left( {c - b}\right) \geq 0 \), so \( z\left( {c - a}\right) \left( {c - b}\right) \geq 0 \) . Moreover,\n\n\[ x\left( {a - c}\right) - y\left( {b - c}\right) \geq x\left... | Yes |
Let \( a, b, c \) be three positive real numbers. Prove that\n\n\[ a + b + c \leq \frac{{a}^{2} + {bc}}{b + c} + \frac{{b}^{2} + {ca}}{c + a} + \frac{{c}^{2} + {ab}}{a + b}. \]\n\n(Ho Joo Lee) | SOLUTION. According to the identity\n\n\[ \frac{{a}^{2} + {bc}}{b + c} - a = \frac{\left( {a - b}\right) \left( {a - c}\right) }{b + c}, \]\n\nwe can change our inequality into the form\n\n\[ x\left( {a - b}\right) \left( {a - c}\right) + y\left( {b - a}\right) \left( {b - c}\right) + z\left( {c - a}\right) \left( {c -... | Yes |
Let \( a, b, c \) be positive real numbers with sum 3 . Prove that\n\n\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{2{a}^{2} + {bc}} + \frac{3}{2{b}^{2} + {ac}} + \frac{3}{2{c}^{2} + {ab}}. \] | SOLUTION. Rewrite the inequality into the following from\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{1}{a} - \mathop{\sum }\limits_{{cyc}}\frac{a + b + c}{2{a}^{2} + {bc}}}\right) \geq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{\left( {a - b}\right) \left( {a - c}\right) }{2{a}^{3} + {abc}} \geq 0. \]\n\... | Yes |
Example 1.1.3. Let \( a, b, c \) be the side lengths of a triangle. Prove that\n\n\[ \frac{\sqrt{a + b - c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} + \frac{\sqrt{b + c - a}}{\sqrt{b} + \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c + a - b}}{\sqrt{c} + \sqrt{a} - \sqrt{b}} \leq 3. \]\n\n(Italian Winter Camp 2007) | Solution. By a simple observation, the inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {1 - \frac{\sqrt{a + b - c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}}}\right) \geq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{\sqrt{a} + \sqrt{b} - \sqrt{c} - \sqrt{a + b - c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} \... | Yes |
Let \( x, y, z \) be positive real numbers such that \( \sqrt{x} + \sqrt{y} + \sqrt{z} = 1 \) . Prove that \[ \frac{{x}^{2} + {yz}}{x\sqrt{2\left( {y + z}\right) }} + \frac{{y}^{2} + {zx}}{y\sqrt{2\left( {z + x}\right) }} + \frac{{z}^{2} + {xy}}{z\sqrt{2\left( {x + y}\right) }} \geq 1. \] | SOLUTION. We use the following simple transformation \[ \mathop{\sum }\limits_{{cyc}}\frac{{x}^{2} + {yz}}{x\sqrt{2\left( {y + z}\right) }} = \mathop{\sum }\limits_{{cyc}}\frac{\left( {x - y}\right) \left( {x - z}\right) + x\left( {y + z}\right) }{x\sqrt{2\left( {y + z}\right) }} \] \[ = \mathop{\sum }\limits_{{cyc}}\f... | Yes |
Example 1.1.5. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{2} + {2bc}}{{\left( b + c\right) }^{2}} + \frac{{b}^{2} + {2ac}}{{\left( a + c\right) }^{2}} + \frac{{c}^{2} + {2ab}}{{\left( a + b\right) }^{2}} \geq \frac{9}{4}. \] | SOLUTION. The inequality can be rewritten as\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{\left( {a - b}\right) \left( {a - c}\right) + \left( {{ab} + {bc} + {ca}}\right) }{{\left( b + c\right) }^{2}} \geq \frac{9}{4}, \]\n\nor equivalently \( A + B \geq \frac{9}{4} \), where\n\n\[ A = \frac{\left( {a - b}\right) \left( {a... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \sqrt{\frac{{a}^{3} + {abc}}{{\left( b + c\right) }^{3}}} + \sqrt{\frac{{b}^{3} + {abc}}{{\left( c + a\right) }^{3}}} + \sqrt{\frac{{c}^{3} + {abc}}{{\left( a + b\right) }^{3}}} \geq \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}. \]\n\n(Nguyen Van Thach... | SOLUTION. Notice that\n\n\[ \sqrt{\frac{{a}^{3} + {abc}}{{\left( b + c\right) }^{3}}} - \frac{a}{b + c} = \frac{\sqrt{a}}{b + c} \cdot \left( {\sqrt{\frac{{a}^{2} + {bc}}{b + c}} - \sqrt{a}}\right) \]\n\n\[ = \frac{\sqrt{a}\left( {a - b}\right) \left( {a - c}\right) }{\left( {b + c}\right) \sqrt{b + c}\left( {\sqrt{{a}... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{2}}{\left( {{2a} + b}\right) \left( {{2a} + c}\right) } + \frac{{b}^{2}}{\left( {{2b} + c}\right) \left( {{2b} + a}\right) } + \frac{{c}^{2}}{\left( {{2c} + a}\right) \left( {{2c} + b}\right) } \leq \frac{1}{3}. \]\n | Solution. If \( c = 0 \), the problem is obvious. Suppose that \( a, b, c > 0 \), then we have\n\n\[ 1 - 3\mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{\left( {{2a} + b}\right) \left( {{2a} + c}\right) } = \mathop{\sum }\limits_{{cyc}}\left( {\frac{a}{a + b + c} - \frac{{a}^{2}}{\left( {{2a} + b}\right) \left( {{2a} + c}... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{{a}^{2} + {2bc}} + \frac{1}{{b}^{2} + {2ca}} + \frac{1}{{c}^{2} + {2ab}} \leq {\left( \frac{a + b + c}{{ab} + {bc} + {ca}}\right) }^{2}. \]\n\n(Pham Huu Duc) | Solution. We have\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{ab} + {bc} + {ca}}{{a}^{2} + {2bc}} - \frac{{\left( a + b + c\right) }^{2}}{{ab} + {bc} + {ca}} = \mathop{\sum }\limits_{{cyc}}\left( {\frac{{ab} + {bc} + {ca}}{{a}^{2} + {2bc}} - 1}\right) + \mathop{\sum }\limits_{{cyc}}\frac{\left( {c - a}\right) \left( {c -... | Yes |
Example 1.1.9. Let \( a, b, c \) be non-negative real numbers such that \( {ab} + {bc} + {ca} = 1 \) . Prove that \[ \frac{1 + {b}^{2}{c}^{2}}{{\left( b + c\right) }^{2}} + \frac{1 + {c}^{2}{a}^{2}}{{\left( c + a\right) }^{2}} + \frac{1 + {a}^{2}{b}^{2}}{{\left( a + b\right) }^{2}} \geq \frac{5}{2}. \] (Mathlinks Conte... | SOLUTION. First we have that \[ \mathop{\sum }\limits_{{cyc}}\frac{1 + {b}^{2}{c}^{2}}{{\left( b + c\right) }^{2}} = \mathop{\sum }\limits_{{cyc}}\frac{{\left( ab + bc + ca\right) }^{2} + {b}^{2}{c}^{2}}{{\left( b + c\right) }^{2}} = \mathop{\sum }\limits_{{cyc}}{a}^{2} + 2\mathop{\sum }\limits_{{cyc}}\frac{abc}{b + c}... | Yes |
Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 1 \) . Prove that\n\n\[ a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) + b\left( {b - a}\right) \left( {b - c}\right) \left( {b - d}\right) + c\left( {c - a}\right) \left( {c - b}\right) \left( {c - d}\right) + d\left( {d... | SOLUTION. We use the entirely mixing variable and the renewed derivative to solve this problem. Notice that our inequality is exactly\n\n\[ \frac{1}{432}{\left( a + b + c + d\right) }^{4} + \mathop{\sum }\limits_{{cyc}}a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) \geq 0. \]\n\nNotice that the ineq... | Yes |
Example 1.2.2. Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) + b\left( {b - a}\right) \left( {b - c}\right) \left( {b - d}\right) + c\left( {c - a}\right) \left( {c - b}\right) \left( {c - d}\right) + d\left( {d - a}\right) \left( ... | SOLUTION. We use the global derivative as in the previous solution. Notice that this inequality is obvious due to Schur inequality if one of four numbers \( a, b, c, d \) is equal to 0 . By taking the global derivative of the left-hand side expression, we only need to prove that\n\n\[ \mathop{\sum }\limits_{{cyc}}\left... | Yes |
Let \( a, b, c, d \) be non-negative real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = 4 \) . Prove that\n\n\[ a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) + b\left( {b - a}\right) \left( {b - c}\right) \left( {b - d}\right) + c\left( {c - a}\right) \left( {c - b}\right) \left( {c ... | We have to prove that\n\n\[ {16}\sum a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) + {\left( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\right) }^{2} - {16abcd} \geq 0. \]\n\nIf \( d = 0 \), the inequality is obvious due to AM-GM inequality and Schur inequality (for three numbers). According to the mixin... | Yes |
Example 1.2.4. Let \( a, b, c, d, e \) be non-negative real numbers such that \( a + b + c + d + e = 1 \) . Prove that\n\n\[ a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) \left( {a - e}\right) + b\left( {b - a}\right) \left( {b - c}\right) \left( {b - d}\right) \left( {b - e}\right) + c\left( {c - ... | SOLUTION. To prove this problem, we have to use two of the previous results. Our inequality is equivalent to\n\n\[ \frac{1}{4320}{\left( a + b + c + d + e\right) }^{5} + \mathop{\sum }\limits_{{cyc}}a\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) \left( {a - e}\right) \geq 0. \]\n\nTaking the global ... | Yes |
Example 1.2.5. Let \( a, b, c, d, e \) be non-negative real numbers. Prove that\n\n\[ \na\left( {a - b}\right) \left( {a - c}\right) \ldots \left( {a - e}\right) + b\left( {b - a}\right) \left( {b - c}\right) \ldots \left( {b - e}\right) + c\left( {c - a}\right) \left( {c - b}\right) \left( {c - d}\right) \left( {c - e... | SOLUTION. This problem is easier than the previous problem. Taking the global derivative for a first time, we obtain an obvious inequality\n\n\[ \n\mathop{\sum }\limits_{{cyc}}\left( {a - b}\right) \left( {a - c}\right) \left( {a - d}\right) \left( {a - e}\right) + 2\mathop{\sum }\limits_{{cyc}}{abcd} + \mathop{\sum }\... | Yes |
Let \( a, b, c, k \) be positive real numbers. Prove that\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{a + k}{b + k} + \frac{b + k}{c + k} + \frac{c + k}{a + k}. \]\n | Solution. Notice that we can transform the expression \( G\left( {a, b, c}\right) \) into\n\n\[ G\left( {a, b, c}\right) = \left( {\frac{a}{b} + \frac{b}{a} - 2}\right) + \left( {\frac{b}{c} + \frac{c}{a} - \frac{b}{a} - 1}\right) = \frac{{\left( a - b\right) }^{2}}{ab} + \frac{\left( {a - c}\right) \left( {b - c}\righ... | Yes |
Let \( a, b, c \) be positive real numbers. If \( k \geq \max \left( {{a}^{2},{b}^{2},{c}^{2}}\right) \), prove that\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{{a}^{2} + k}{{b}^{2} + k} + \frac{{b}^{2} + k}{{c}^{2} + k} + \frac{{c}^{2} + k}{{a}^{2} + k}. \] | Solution. Similarly as in the preceding inequality, this one is equivalent to\n\n\[ \frac{{\left( a - b\right) }^{2}}{ab} + \frac{\left( {a - c}\right) \left( {b - c}\right) }{ac} \geq \frac{{\left( a - b\right) }^{2}{\left( a + b\right) }^{2}}{\left( {{a}^{2} + k}\right) \left( {{b}^{2} + k}\right) } + \frac{\left( {a... | Yes |
If \( a, b, c \) are the side lengths of a triangle, then\n\n\[ 4\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}\right) \geq 9 + \frac{{a}^{2} + {c}^{2}}{{c}^{2} + {b}^{2}} + \frac{{c}^{2} + {b}^{2}}{{b}^{2} + {a}^{2}} + \frac{{b}^{2} + {a}^{2}}{{a}^{2} + {c}^{2}}. \] | The inequality can be rewritten in the following form\n\n\[ \frac{4{\left( a - b\right) }^{2}}{ab} + \frac{4\left( {c - a}\right) \left( {c - b}\right) }{ac} \geq \frac{{\left( {a}^{2} - {b}^{2}\right) }^{2}}{\left( {{a}^{2} + {c}^{2}}\right) \left( {{c}^{2} + {b}^{2}}\right) } + \frac{\left( {{c}^{2} - {a}^{2}}\right)... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{2}}{{b}^{2}} + \frac{{b}^{2}}{{c}^{2}} + \frac{{c}^{2}}{{a}^{2}} + \frac{8\left( {{ab} + {bc} + {ca}}\right) }{{a}^{2} + {b}^{2} + {c}^{2}} \geq {11}. \]\n\n(Nguyen Van Thach) | SOLUTION. Similarly, this inequality can be rewritten in the following form\n\n\[ {\left( a - b\right) }^{2}\left( {\frac{{\left( a + b\right) }^{2}}{{a}^{2}{b}^{2}} - \frac{8}{{a}^{2} + {b}^{2} + {c}^{2}}}\right) + \left( {c - a}\right) \left( {c - b}\right) \left( {\frac{\left( {a + c}\right) \left( {b + c}\right) }{... | Yes |
Let \( a, b, c \) be the side lengths of a triangle. Prove that\n\n\[ \frac{{a}^{2} + {b}^{2}}{{a}^{2} + {c}^{2}} + \frac{{c}^{2} + {a}^{2}}{{c}^{2} + {b}^{2}} + \frac{{b}^{2} + {c}^{2}}{{b}^{2} + {a}^{2}} \geq \frac{a + b}{a + c} + \frac{a + c}{b + c} + \frac{b + c}{b + a}. \]\n\n(Vo Quoc Ba Can) | SOLUTION. It is easy to rewrite the inequality in the following form\n\n\[ {\left( a - b\right) }^{2}M + \left( {c - a}\right) \left( {c - b}\right) N \geq 0, \]\n\nwhere\n\n\[ M = \frac{{\left( a + b\right) }^{2}}{\left( {{a}^{2} + {c}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) } - \frac{1}{\left( {a + c}\right) \l... | Yes |
For all distinct real numbers \( a, b, c \), prove that\n\n\[ \frac{{\left( a - b\right) }^{2}}{{\left( b - c\right) }^{2}} + \frac{{\left( b - c\right) }^{2}}{{\left( c - a\right) }^{2}} + \frac{{\left( c - a\right) }^{2}}{{\left( a - b\right) }^{2}} \geq 5. \]\n\n(Darij Grinberg) | This inequality is directly deduced from the following identity\n\n\[ \frac{{\left( a - b\right) }^{2}}{{\left( b - c\right) }^{2}} + \frac{{\left( b - c\right) }^{2}}{{\left( c - a\right) }^{2}} + \frac{{\left( c - a\right) }^{2}}{{\left( a - b\right) }^{2}} = 5 + {\left( 1 + \frac{a - b}{b - c} + \frac{b - c}{c - a} ... | Yes |
Prove that for all positive real numbers \( a, b, c \), we have\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \sqrt{\frac{{a}^{2} + {c}^{2}}{{b}^{2} + {c}^{2}}} + \sqrt{\frac{{c}^{2} + {b}^{2}}{{a}^{2} + {b}^{2}}} + \sqrt{\frac{{b}^{2} + {a}^{2}}{{c}^{2} + {a}^{2}}} \] | SOLUTION. First Solution. First, we will prove that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a}{b} + \mathop{\sum }\limits_{{cyc}}\frac{b}{a} \geq \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{{a}^{2} + {c}^{2}}{{b}^{2} + {c}^{2}}} + \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{{b}^{2} + {c}^{2}}{{a}^{2} + {c}^{2}}} \]\n\n(1)\n... | Yes |
Let \( a, b, c \) be the side lengths of a triangle. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{3}{2} \leq \frac{2ab}{c\left( {a + b}\right) } + \frac{2bc}{a\left( {b + c}\right) } + \frac{2ca}{b\left( {c + a}\right) }.\]\n\nor, in other words, prove that \( N\left( {a, b, c}\right) \l... | SOLUTION. First, we change the inequality to SOS form as follows\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {2{c}^{2} - {ab}}\right) \left( {a + b}\right) {\left( a - b\right) }^{2} \geq 0.\]\n\nWLOG, assume that \( a \geq b \geq c \), then \( {S}_{a} \geq {S}_{b} \geq {S}_{c} \) . Therefore, it’s enough to prove that\n... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{2ab}{c\left( {a + b}\right) } + \frac{2bc}{a\left( {b + c}\right) } + \frac{2ca}{b\left( {c + a}\right) } \geq \frac{a + b}{{2c} + a + b} + \frac{b + c}{{2a} + b + c} + \frac{c + a}{{2b} + c + a}, \]\n\nor, in other words, prove that \( N\left( {\frac{1... | SOLUTION. Similarly to the preceding problem, after changing the inequality to SOS form, we only need to prove that\n\n\[ \frac{a\left( {{b}^{3} + {c}^{3}}\right) + {bc}\left( {{b}^{2} + {c}^{2}}\right) }{{abc}\mathop{\prod }\limits_{{cyc}}\left( {a + b}\right) } \geq \frac{{2a} + {3b} + {3c}}{\mathop{\prod }\limits_{{... | Yes |
Let \( a, b, c \) be the side lengths of a triangle. Prove that\n\n\[ F\left( {a, b, c}\right) \leq F\left( {a + b, b + c, c + a}\right) . \] | Solution. Notice that the expression \( F\left( {a, b, c}\right) \) can be rewritten as\n\n\[ F\left( {a, b, c}\right) = \mathop{\sum }\limits_{{cyc}}a\left( {a - b}\right) \left( {a - c}\right) . \]\n\nTherefore our inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {b + c - a}\right) \left( {a - b}... | Yes |
Let \( a, b, c \) be the side lengths of a triangle. Prove that\n\n\[ F\left( {a, b, c}\right) \leq 4{a}^{2}{b}^{2}{c}^{2}F\left( {\frac{1}{a},\frac{1}{b},\frac{1}{c}}\right) . \]\n\n(Pham Kim Hung) | Solution. Generally, the expression \( F\left( {a, b, c}\right) \) can be represented in SOS form as\n\n\[ F\left( {a, b, c}\right) = \frac{1}{2}\mathop{\sum }\limits_{{cyc}}\left( {a + b - c}\right) {\left( a - b\right) }^{2}. \]\n\nTherefore we can change our inequality to the following\n\n\[ \mathop{\sum }\limits_{{... | Yes |
Prove that if \( a, b, c \) are the side lengths of an acute triangle then\n\n\[ {3F}\left( {a, b, c}\right) \geq F\left( {a - b, b - c, c - a}\right) . \] | SOLUTION. We will only prove the second part of this problem because the first part can be deduced similarly but simpler. Now suppose that \( a, b, c \) are side lengths of an acute triangle. Clearly, if \( x + y + z = 0 \) then\n\n\[ {x}^{3} + {y}^{3} + {z}^{3} + {3xyz} - {xy}\left( {y + x}\right) - {yz}\left( {y + z}... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{9F}\left( {{a}^{2},{b}^{2},{c}^{2}}\right) \geq {8F}\left( {{\left( a - b\right) }^{2},{\left( b - c\right) }^{2},{\left( c - a\right) }^{2}}\right) .\n\]\n(Pham Kim Hung) | SOLUTION. We use the mixing all variables method, similarly as in the preceding problem. We can assume that \( a \geq b \geq c = 0 \) . In this case, we obtain\n\n\[ \nF\left( {{\left( a - b\right) }^{2},{\left( b - c\right) }^{2},{\left( c - a\right) }^{2}}\right) = {a}^{6} + {b}^{6} + {\left( a - b\right) }^{6} + 3{a... | Yes |
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c = 3 \) . Find the maximum of the following expressions\n\n(a) \( {S}_{2} = {2}^{ab} + {2}^{bc} + {2}^{ca} \) .\n\n(b) \( {S}_{4} = {4}^{ab} + {4}^{bc} + {4}^{ca} \) . | SOLUTION. Don’t hurry to conclude that \( \max {S}_{2} = 6 \) and \( \max {S}_{4} = {12} \) because the reality is different. We figure out a solution by the mixing variable method and solve a general problem that involves both \( \left( a\right) \) and \( \left( b\right) \) . WLOG, assume that \( a \geq b \geq c \) an... | Yes |
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c = 3 \) . Prove that\n\n\[ \n{2}^{4ab} + {2}^{4bc} + {2}^{4ca} - {2}^{3abc} \leq {513}.\n\]\n\n(Pham Kim Hung) | SOLUTION. In fact, this problem reminds us of Schur inequality only that we know have exponents (notice that if \( a + b + c = 3 \) then Schur inequality is equivalent to \( 4\left( {{ab} + {bc} + {ca}}\right) \leq 9 + {3abc}) \) . According to the example 1.6.1, we deduce that\n\n\[ \n{2}^{4ab} + {2}^{4bc} + {2}^{4ca}... | Yes |
Let \( x, y, z \) be non-negative real numbers. Prove that\n\n\[ \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} + \frac{{25}\left( {{xy} + {yz} + {zx}}\right) }{{\left( x + y + z\right) }^{2}} \geq 8. \] | Solution. WLOG, assume that \( x \geq y \geq z \) . Denote\n\n\[ f\left( {x, y, z}\right) = \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} + \frac{{25}\left( {{xy} + {yz} + {zx}}\right) }{{\left( x + y + z\right) }^{2}}. \]\n\nWe infer that\n\n\[ f\left( {x, y, z}\right) - f\left( {x, y + z,0}\right) = \frac{y}{z ... | Yes |
Let \( x, y, z \) be non-negative real numbers. Prove that\n\n\[ \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} + \frac{{16}\left( {{xy} + {yz} + {zx}}\right) }{{x}^{2} + {y}^{2} + {z}^{2}} \geq 8. \] | We use mixing variables to solve this problem. Denote \( x = \max \{ x, y, z\} \) and\n\n\[ f\left( {x, y, z}\right) = \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} + \frac{{16}\left( {{xy} + {yz} + {zx}}\right) }{{x}^{2} + {y}^{2} + {z}^{2}}. \]\n\nIt is easy to see that the statament \( f\left( {x, y, z}\right)... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 1. Prove that\n\n\[ \frac{{a}^{2}}{{b}^{2} + {c}^{2}} + \frac{{b}^{2}}{{c}^{2} + {a}^{2}} + \frac{{c}^{2}}{{a}^{2} + {b}^{2}} + \frac{{27}{\left( a + b + c\right) }^{2}}{{a}^{2} + {b}^{2} + {c}^{2}} \geq {52}. \]\n\n(Pham Kim Hung) | Solution. WLOG, assume that \( a \geq b \geq c \) . Denote\n\n\[ f\left( {a, b, c}\right) = \frac{{a}^{2}}{{b}^{2} + {c}^{2}} + \frac{{b}^{2}}{{c}^{2} + {a}^{2}} + \frac{{c}^{2}}{{a}^{2} + {b}^{2}} + \frac{{27}{\left( a + b + c\right) }^{2}}{{a}^{2} + {b}^{2} + {c}^{2}}. \]\n\nWe will prove that \( f\left( {a, b, c}\ri... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{{\left( a + b\right) }^{2}} + \frac{1}{{\left( b + c\right) }^{2}} + \frac{1}{{\left( c + a\right) }^{2}} + \frac{24}{{\left( a + b + c\right) }^{2}} \geq \frac{8}{{ab} + {bc} + {ca}}. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . Denote\n\n\[ f\left( {a, b, c}\right) = \frac{1}{{\left( a + b\right) }^{2}} + \frac{1}{{\left( b + c\right) }^{2}} + \frac{1}{{\left( c + a\right) }^{2}} + \frac{24}{{\left( a + b + c\right) }^{2}} - \frac{8}{{ab} + {bc} + {ca}}. \]\n\nWe get that\n\n\[ f\left( {a, b... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{4\left( {a + b + c}\right) \left( {{ab} + {bc} + {ca}}\right) }{{a}^{3} + {b}^{3} + {c}^{3}} \geq 5. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . Denote \( t = b + c \) and\n\n\[ f\left( {a, b, c}\right) = \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{4\left( {a + b + c}\right) \left( {{ab} + {bc} + {ca}}\right) }{{a}^{3} + {b}^{3} + {c}^{3}}. \]\n\nWe have\n\n\[ f\left( {a, b, c}\right) - f\left(... | Yes |
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