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Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{a}{b + c}} + \sqrt{\frac{b}{c + a}} + \sqrt{\frac{c}{a + b}} + \frac{9\sqrt{{ab} + {bc} + {ca}}}{a + b + c} \geq 6. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . We have\n\n\[ \sqrt{\frac{ab}{a + c}} + \sqrt{\frac{ac}{a + b}} \geq \sqrt{\frac{b \cdot b}{b + c}} + \sqrt{\frac{c \cdot c}{c + b}} = \sqrt{b + c} \]\n\n\[ \Rightarrow \sqrt{\frac{b}{c + a}} + \sqrt{\frac{c}{a + b}} \geq \sqrt{\frac{b + c}{a}}. \]\n\nLet now \( t = b... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{a}{b + c}} + \sqrt{\frac{b}{c + a}} + \sqrt{\frac{c}{a + b}} + \sqrt{\frac{{27}\left( {{ab} + {bc} + {ca}}\right) }{{a}^{2} + {b}^{2} + {c}^{2}}} \geq \frac{7\sqrt{2}}{2}. \]\n\n(Vo Quoc Ba Can) | SOLUTION. Similarly as in the previous problem, we get that if \( a \geq b \geq c \) and \( t = b + c \) then\n\n\[ \sqrt{\frac{a}{b + c}} + \sqrt{\frac{b}{c + a}} + \sqrt{\frac{c}{a + b}} + \sqrt{\frac{{27}\left( {{ab} + {bc} + {ca}}\right) }{{a}^{2} + {b}^{2} + {c}^{2}}} \geq \frac{a + t}{\sqrt{at}} + 3\sqrt{3} \cdot... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} + \frac{8}{a + b + c} \geq \frac{6}{\sqrt{{ab} + {bc} + {ca}}}. \]\n\n(Pham Kim Hung) | Solution. Similarly to the previous proofs, we assume first that \( a \geq b \geq c \) and denote\n\n\[ f\left( {a, b, c}\right) = \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} + \frac{8}{a + b + c} - \frac{6}{\sqrt{{ab} + {bc} + {ca}}}. \]\n\nLet \( t = b + c \), then we have\n\n\[ f\left( {a, b, c}\right) - f\l... | Yes |
Suppose that \( a, b, c \) are three positive real numbers satisfying\n\n\[ \left( {a + b + c}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\right) = {13}. \]\n\nFind the minimum value of\n\n\[ P = \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \left( {\frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{c}^{2}}}\r... | SOLUTION. Let now\n\[ x = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}, y = \frac{b}{a} + \frac{c}{b} + \frac{a}{c}, \]\n\nthen we obtain that\n\n\[ {x}^{2} = {2y} + \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{{b}^{2}};{y}^{2} = {2x} + \mathop{\sum }\limits_{{cyc}}\frac{{b}^{2}}{{a}^{2}}; \]\n\nBecause \( x + y = {10} \), A... | Yes |
Let \( a, b, c \) be three positive real numbers such that\n\n\[ \left( {a + b + c}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\right) = {16}.\n\]\n\nFind the maximum value of\n\n\[ P = \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \left( {\frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{c}^{2}}}\right) . | SOLUTION. First Solution. Taking into account an example in Volume I, we deduce\n\nthat\n\n\[ \frac{{13} - \sqrt{5}}{2} \leq \mathop{\sum }\limits_{{cyc}}\frac{a}{b} \leq \frac{{13} + \sqrt{5}}{2} \]\n\n\[ \frac{{13} - \sqrt{5}}{2} \leq \mathop{\sum }\limits_{{cyc}}\frac{b}{a} \leq \frac{{13} + \sqrt{5}}{2}. \]\n\nther... | Yes |
Suppose that \( a, b, c \) are three positive real numbers satisfying that\n\n\[ \left( {a + b + c}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\right) = {16}. \]\n\nFind the minimum and maximum value of\n\n\[ P = \left( {{a}^{4} + {b}^{4} + {c}^{4}}\right) \left( {\frac{1}{{a}^{4}} + \frac{1}{{b}^{4}} + \fr... | Solution. We denote \( x, y, m, n \) as in the previous problem and also denote\n\n\[ p = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{{b}^{2}};q = \mathop{\sum }\limits_{{cyc}}\frac{{b}^{2}}{{a}^{2}} \]\n\nThe expression \( P \) can be rewritten as\n\n\[ P = 3 + \mathop{\sum }\limits_{{cyc}}\frac{{a}^{4}}{{b}^{4}} = 3 ... | Yes |
Let \( a, b, c, d \) be positive real numbers such that\n\n\[ \left( {a + b + c + d}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}\right) = {20}. \]\n\nProve that\n\n\[ \left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) \left( {\frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{c}^{2}} + \frac... | SOLUTION. For each triple of positive real numbers \( \left( {x, y, z}\right) \), we denote\n\n\[ F\left( {x, y, z}\right) = \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \]\n\nClearly\n\n\[ F{\left( x, y, z\right) }^{2} = F\left( {{x}^{2},{y}^{2},{z}^{2}}\right) + {2F}\left( {z, y, x}\right) ;F{\left( z, y, x\right) }^{2} =... | Yes |
Example 1.8.5. Suppose that \( a, b, c \) are three positive real numbers satisfying\n\n\[ \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) {\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) }^{2} = {36}. \]\n\nProve the following inequality\n\n\[ {\left( a + b + c\right) }^{2}\left( {\frac{1}{{a}^{2}} + \frac{1}{{b}^{2... | SOLUTION. As in the preceding problems, we denote\n\n\[ x = \mathop{\sum }\limits_{{cyc}}\frac{a}{b};y = \mathop{\sum }\limits_{{cyc}}\frac{b}{a};m = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{bc};n = \mathop{\sum }\limits_{{cyc}}\frac{bc}{{a}^{2}}; \]\n\nThe hypothesis shows that\n\n\[ \mathop{\sum }\limits_{{cyc}}\f... | Yes |
Theorem 3 (CD3). Let \( P\\left( {a, b, c}\\right) \) be a cyclic homogeneous polynomial of degree 3 . The inequality \( P \\geq 0 \) holds for all non-negative variables \( a, b, c \) if and only if\n\n\[ P\\left( {1,1,1}\\right) \\geq 0;P\\left( {a, b,0}\\right) \\geq 0\\forall a, b \\geq 0; \] | Proof. The necessary condition is obvious. We only need to consider the sufficient condition. Assume that\n\n\[ P\\left( {1,1,1}\\right) \\geq 0;P\\left( {a, b,0}\\right) \\geq 0\\forall a, b \\geq 0; \]\n\nWe will prove that for all \( a, b, c \\geq 0 \) we have\n\n\[ P\\left( {a, b, c}\\right) = m\\mathop{\\sum }\\li... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + \left( {\\frac{3}{\\sqrt[3]{4}} - 1}\\right) {abc} \\geq \\frac{3}{\\sqrt[3]{4}}\\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\\right) .\n\] | Solution. Clearly, the above inequality is true if \( a = b = c \) . According to CD3 theorem, it suffices to consider the inequality in one case \( b = 1, c = 0 \) . The inequality becomes\n\n\[ \n{a}^{3} + 1 \\geq \\frac{3}{\\sqrt[3]{4}}{a}^{2}\n\]\n\nwhich simply follows from AM-GM inequality\n\n\[ \n{a}^{3} + 1 = \... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 3. Prove that\n\n\[ \n{a}^{2}b + {b}^{2}c + {c}^{2}a + {abc} \leq 4.\n\] | SOLUTION. The inequality is equivalent to\n\n\[ \n{27}\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a + {abc}}\right) \leq 4{\left( a + b + c\right) }^{3}.\n\]\n\nBecause this is a cyclic inequality and holds for \( a = b = c \), we only need to consider the case \( c = 0 \) due to CD3 theorem. In this case, the inequality beco... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ 4\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) + {12}\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) \geq {15}\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) + {3abc}. \]\n | SOLUTION. Because the inequality is cyclic and holds for \( a = b = c \), according to CD3 theorem, it is enough to consider the case \( c = 0 \) . The inequality becomes\n\n\[ 4\left( {{a}^{3} + {b}^{3}}\right) + {12}{a}^{2}b \geq {15a}{b}^{2}, \]\n\nor\n\n\[ {\left( 2a - b\right) }^{2}\left( {a + {4b}}\right) \geq 0,... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 4. Prove that\n\n\[ 5\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) + {3abc} \leq {36} + 3\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) . \] | SOLUTION. Because the inequality is cyclic and holds for \( a = b = c = \frac{4}{3} \), it suffices to consider it in case \( c = 0 \) and \( a + b = 4 \) . We have to prove that\n\n\[ {5a}{b}^{2} - 3{a}^{2}b \leq {36} \]\n\nor\n\n\[ a\left( {4 - a}\right) \left( {5 - {2a}}\right) \leq 9. \]\n\nApplying AM-GM inequalit... | Yes |
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c = 3 \) . For each \( k \geq 0 \), find the maximum value of\n\n\[ \n{a}^{2}\left( {{kb} + c}\right) + {b}^{2}\left( {{kc} + a}\right) + {c}^{2}\left( {{ka} + b}\right) .\n\]\n\n(Pham Kim Hung) | SOLUTION. Because the expression is cyclic, we can assume first that \( c = 0, a + b = 3 \) and find the maximum value of\n\n\[ \nF = k{a}^{2}b + {b}^{2}a.\n\]\n\nFor \( k = 1 \), we have \( F = {ab}\left( {a + b}\right) = {3ab} \leq \frac{27}{4} \) by AM-GM inequality. Otherwise, assume that \( k \neq 1 \) . We denote... | Yes |
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c = 3 \) . For each \( k \geq 0 \), find the maximum and minimum value of\n\n\[ \n{a}^{2}\left( {{kb} - c}\right) + {b}^{2}\left( {{kc} - a}\right) + {c}^{2}\left( {{ka} - b}\right) .\n\]\n\n(Pham Kim Hung) | SOLUTION. As in the preceding solutions, we will first consider the case \( c = 0 \) . Denote\n\n\[ \nk{a}^{2}b - a{b}^{2} = k{a}^{2}\left( {3 - a}\right) - a{\left( 3 - a\right) }^{2} = - \left( {k + 1}\right) {a}^{3} + 3\left( {k + 2}\right) {a}^{2} - {9a} = f\left( a\right) ,\n\]\n\nthen we get\n\n\[ \n{f}^{\prime }... | Yes |
Theorem 4 (CD3-improved). Let \( P\left( {a, b, c}\right) \) be a cyclic polynomial of degree 3 . \n\n\[ \nP\left( {a, b, c}\right) = m\mathop{\sum }\limits_{{cyc}}{a}^{3} + n\mathop{\sum }\limits_{{cyc}}{a}^{2}b + p\mathop{\sum }\limits_{{cyc}}a{b}^{2} + {qabc} + r\mathop{\sum }\limits_{{cyc}}{a}^{2} + s\mathop{\sum }... | Proof. We fix the sum \( a + b + c = A \) and prove that for all \( A \geq 0 \) then \n\n\[ \nQ\left( {a, b, c}\right) = m\mathop{\sum }\limits_{{cyc}}{a}^{3} + n\mathop{\sum }\limits_{{cyc}}{a}^{2}b + p\mathop{\sum }\limits_{{cyc}}a{b}^{2} + {qabc} + \frac{r}{A}\left( {\mathop{\sum }\limits_{{cyc}}{a}^{2} + s\mathop{\... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + 2 + \frac{4}{3}\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) \geq 3\left( {{ab} + {bc} + {ca}}\right) .\n\]\n\n(Pham Kim Hung) | SOLUTION. For \( a = b = c = t \), the inequality becomes\n\n\[ \n4{t}^{3} - 6{t}^{2} + 2 \geq 0 \Leftrightarrow 2{\left( t - 1\right) }^{2}\left( {{2t} + 1}\right) \geq 0.\n\]\n\nThis one is obvious, so we are done. Equality holds for \( a = b = c = 1 \).\n\nFor \( c = 0 \), the inequality becomes\n\n\[ \n\frac{4}{3}{... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + 2\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) + {12} \geq 6\left( {a + b + c}\right) + {ab} + {bc} + {ca}. \n\]\n\n(Pham Kim Hung) | Solution. For \( a = b = c = t \), the inequality becomes \( 6{t}^{3} + {12} \geq {18t} \), or \( 6{\left( t - 1\right) }^{2}(t + \) \( 2) \geq 0 \), which is obvious. Therefore it suffices to prove the inequality in case \( c = 0 \) . In this case, we have to prove that\n\n\[ \n{a}^{2} + {b}^{2} + 2{a}^{2}b + {12} \ge... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + 3 + \frac{1}{6}\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a + {15abc}}\right) \geq a + b + c + 2\left( {{ab} + {bc} + {ca}}\right) .\n\]\n\n(Pham Kim Hung) | Solution. If \( a = b = c \), the inequality is obvious. If \( c = 0 \), it becomes\n\n\[ \n{a}^{2} + {b}^{2} + 3 + \frac{{a}^{2}b}{6} \geq a + b + {2ab}\n\]\n\nor\n\[ \n{a}^{2}\left( {1 + \frac{b}{6}}\right) - \left( {{2b} + 1}\right) a + \left( {{b}^{2} - b + 3}\right) \geq 0.\n\]\n\nIt is easy to check that\n\n\[ \n... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[\n2\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) + {abc} + {ab} + {bc} + {ca} + 2 \geq 2\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) + {a}^{2} + {b}^{2} + {c}^{2} + a + b + c.\n\] | Solution. If \( a = b = c \), the inequality is obvious, with equality for \( a = b = c = 1 \) . Therefore we can assume that \( c = 0 \), and the inequality becomes\n\n\[\n2\left( {{a}^{3} + {b}^{3}}\right) + {ab} + 2 \geq 2{a}^{2}b + {a}^{2} + {b}^{2} + a + b.\n\]\n\nWe may assume that \( a \geq b \) . Denote\n\n\[\n... | Yes |
Proposition 1. Let \( P\left( {a, b, c, d}\right) \) be a symmetric polynomial of degree 3 . The inequality \( P \geq \) 0 holds for all non-negative variables \( a, b, c, d \) if and only if\n\n\[ P\left( {1,1,1,1}\right) \geq 0;P\left( {1,1,1,0}\right) \geq 0;P\left( {1,1,0,0}\right) \geq 0;P\left( {1,0,0,0}\right) \... | SOLUTION. The necessary condition is obvious. To prove the sufficient condition, we will use Hoo Joo Lee's theorem. WLOG, assume that\n\n\[ P\left( {a, b, c, d}\right) = \alpha \mathop{\sum }\limits_{{cyc}}{a}^{3} + \beta \mathop{\sum }\limits_{{cyc}}{a}^{2}\left( {b + c + d}\right) + {6\gamma }\mathop{\sum }\limits_{{... | Yes |
Proposition 2. Let \( P\\left( {{x}_{1},{x}_{2},\\ldots ,{x}_{n}}\\right) \) be a third-degree symmetric polynomial\n\n\[ P = \\alpha \\mathop{\\sum }\\limits_{{i = 1}}^{n}{x}_{i}^{3} + \\beta \\mathop{\\sum }\\limits_{{i \\neq j}}{x}_{i}^{2}{x}_{j} + \\gamma \\mathop{\\sum }\\limits_{{i \\neq j \\neq k}}{x}_{i}{x}_{j}... | Proof. To prove this problem, we use induction. Assuming that the theorem is proved already for \( n - 1 \) variables, we have to prove it for \( n \) variables. Generally, the polynomial \( P\\left( {{x}_{1},{x}_{2},\\ldots ,{x}_{n}}\\right) \) can be expressed in the following form\n\n\[ P = \\alpha \\mathop{\\sum }\... | Yes |
Example 1.10.11. Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ 4\left( {{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3}}\right) + {15}\left( {{abc} + {bcd} + {cda} + {dab}}\right) \geq {\left( a + b + c + d\right) }^{3}. \]\n | SOLUTION. Because this is a third-degree symmetric inequality of four variables, according to the generalization of the SD3 theorem, it suffices to check this inequality in case \( a = b = c = d = 1 \) or \( a = 0, b = c = d = 1 \) or \( a = b = 0, c = d = 1 \) or \( a = b = c = 0, d = 1 \) . They are all obvious, so w... | Yes |
Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3} + {10}\left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) \leq {64}.\n\] | Solution. The inequality can be rewritten as (homogeneous form)\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3} + \frac{5}{2}\left( {a + b + c + d}\right) \left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) \leq {\left( a + b + c + d\right) }^{3}.\n\]\n\nIt is easy to check that the inequality holds for \( \left( {a,... | No |
Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3} + \frac{14}{3}\left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) \geq {32}.\n\] | Solution. The inequality can be rewritten as (homogeneous form)\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3} + \frac{7}{6}\left( {a + b + c + d}\right) \left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) \geq \frac{1}{2}{\left( a + b + c + d\right) }^{3}.\n\]\n\nBy the previous proposition/theorem, it suffices to ... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n}\left( {n \geq 3}\right) \) be non-negative real numbers. Prove that\n\n\[ \frac{n - 1}{2}\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{3} + \frac{3}{n - 2}\mathop{\sum }\limits_{{cyc}}{a}_{1}{a}_{2}{a}_{3} \geq \mathop{\sum }\limits_{{cyc}}{a}_{1}{a}_{2}\left( {{a}_{1} + {a}_{2}}\ri... | Solution. In this problem, we have \( \alpha = \frac{n - 1}{2} \) and \( \beta = - 1 \) . Because\n\n\[ {3\alpha } + \left( {n - 1}\right) \beta = \frac{3\left( {n - 1}\right) }{2} - \left( {n - 1}\right) = \frac{n - 1}{2} > 0, \]\n\naccording to the previous theorem (generalization for \( n \) variables), we get that ... | Yes |
Proposition 3. Consider the following expression of non-negative real numbers \( a, b, c \)\n\n\[ \nF\left( {a, b, c}\right) = \frac{{ma} + {nb} + {pc}}{{\alpha a} + {\beta b} + {\gamma c}} + \frac{{mb} + {nc} + {pa}}{{\alpha b} + {\beta c} + {\gamma a}} + \frac{{mc} + {na} + {pb}}{{\alpha c} + {\beta a} + {\gamma b}},... | Proof. It is easy to see that this proposition is directly obtained from the theorem CD3. Indeed, consider the expression\n\n\[ \nG\left( {a, b, c}\right) = \mathop{\sum }\limits_{{cyc}}\left( {{ma} + {bp} + {pc}}\right) \left( {{\alpha b} + {\beta c} + {\gamma a}}\right) \left( {{\alpha c} + {\beta a} + {\gamma b}}\ri... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a + {2b}}{c + {2b}} + \frac{b + {2c}}{a + {2c}} + \frac{c + {2a}}{b + {2a}} \geq 3. \] | The inequality is cyclic and holds for \( a = b = c \), so, according to the previous theorem, we can assume that \( b = 1, c = 0 \) . In this case, we have to prove that\n\n\[ \frac{a + 2}{2} + \frac{1}{a} + \frac{2a}{1 + {2a}} \geq 3 \]\n\n\[ \Leftrightarrow \frac{a}{2} + \frac{a + 1}{a\left( {1 + {2a}}\right) } \geq... | No |
Let \( a, b, c \) be non-negative real numbers. For each \( k \geq 0 \), find the minimum of the expression\n\n\[ \frac{a + {kb}}{c + {kb}} + \frac{b + {kc}}{a + {kc}} + \frac{c + {ka}}{b + {ka}}. \] | Solution. For \( b = 1, c = 0 \), the expression becomes\n\n\[ f\left( a\right) = \frac{a + k}{k} + \frac{1}{a} + \frac{ka}{1 + {ka}} = 2 + \frac{a}{k} + \frac{1}{a} - \frac{1}{1 + {ka}}. \]\n\nIf \( k \leq 1 \) then, according to AM-GM inequality, we have\n\n\[ f\left( a\right) \geq 2 + \frac{2}{\sqrt{k}} - 1 \geq 3 \... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ 1 \leq \frac{a + b}{a + {4b} + c} + \frac{b + c}{b + {4c} + a} + \frac{c + a}{c + {4a} + b} \leq \frac{4}{3}. \]\n\n(Pham Kim Hung) | Solution. For \( b = 1, c = 0 \), the inequality becomes\n\n\[ \frac{a + 1}{a + 4} + \frac{1}{a + 1} + \frac{a}{{4a} + 1} \geq 1\left( \star \right) \]\n\nand\n\n\[ \frac{a + 1}{a + 4} + \frac{1}{a + 1} + \frac{a}{{4a} + 1} \leq \frac{4}{3}\left( {\star \star }\right) \]\n\nThe inequality \( \left( \star \right) \) is ... | No |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{a}{{4a} + {4b} + c}} + \sqrt{\frac{b}{{4b} + {4c} + a}} + \sqrt{\frac{c}{{4c} + {4a} + b}} \leq 1. \]\n\n(Pham Kim Hung, Volume I) | SOLUTION. It suffices to prove that\n\n\[ \frac{a}{{4a} + {4b} + c} + \frac{b}{{4b} + {4c} + a} + \frac{c}{{4c} + {4a} + b} \leq \frac{1}{3}\left( \star \right) \]\n\nFor \( c = 0 \), the inequality becomes\n\n\[ \frac{a}{{4a} + {4b}} + \frac{b}{{4b} + a} \leq \frac{1}{3} \]\n\nor (after expanding)\n\n\[ {\left( a - 2b... | No |
Let \( a, b, c \) be non-negative real numbers. For each \( k, l \geq 0 \), find the maximal and minimal value of the expression\n\n\[ \frac{a}{{ka} + {lb} + c} + \frac{b}{{kb} + {lc} + a} + \frac{c}{{kc} + {la} + b}. \] | Solution. First we will examine the expression in case \( b = 1, c = 0 \) . Denote\n\n\[ f\left( a\right) = \frac{a}{{ka} + l} + \frac{1}{k + a} \]\n\nthen we get\n\n\[ {f}^{\prime }\left( a\right) = \frac{l}{{\left( ka + l\right) }^{2}} - \frac{1}{{\left( k + a\right) }^{2}}. \]\n\nThe equation \( {f}^{\prime }\left( ... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n\\frac{{3a} + {2b} + c}{a + {2b} + {3c}} + \\frac{{3b} + {2c} + a}{b + {2c} + {3a}} + \\frac{{3c} + {2a} + b}{c + {2a} + {3b}} \\geq 3.\n\] | SOLUTION. First we have to consider the inequality in case \( b = 1, c = 0 \) . In this case, the inequality becomes\n\n\[ \n\\frac{{3a} + 2}{a + 2} + \\frac{3 + a}{{3a} + 1} + \\frac{{2a} + 1}{{2a} + 3} \\geq 3 \n\]\n\n\[ \n\\Leftrightarrow \\frac{2a}{a + 2} + \\frac{3 + a}{{3a} + 1} \\geq \\frac{{2a} + 5}{{2a} + 3} \... | No |
Let \( a, b, c \) be non-negative real numbers with sum 3. Prove that\n\n\[ \frac{a + b}{1 + b} + \frac{b + c}{1 + c} + \frac{c + a}{1 + a} \geq 3 \] | Solution. The inequality is equivalent to (homogeneous form)\n\n\[ \frac{a + b}{b + \frac{a + b + c}{3}} + \frac{b + c}{c + \frac{a + b + c}{3}} + \frac{c + a}{a + \frac{a + b + c}{3}} \geq 3. \]\n\nBecause this problem is cyclic, homogeneous and holds if \( a = b = c \), we can assume that \( c = 0 \) and \( a + b = 3... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be real numbers. Prove that\n\n\[ \mathop{\sum }\limits_{{i, j = 1}}^{n}\frac{{a}_{i}{a}_{j}}{i + j} \geq 0 \] | SOLUTION. This problem shows the great advantage of the integral method because other solutions are almost impossible. Indeed, consider the following function\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{i, j = 1}}^{n}{a}_{i}{a}_{j}{x}^{i + j - 1} = \frac{1}{x}{\left( \mathop{\sum }\limits_{{i, j = 1}}^{n}{a}_{i}{x... | Yes |
Let \( a, b, c \) be positive real numbers such that \( a + b + c = 1 \) . Prove that\n\n\[ \left( {{ab} + {bc} + {ca}}\right) \left( {\frac{a}{{b}^{2} + b} + \frac{b}{{c}^{2} + c} + \frac{c}{{a}^{2} + a}}\right) \geq \frac{3}{4}. \] | Solution. We will prove first the following result: for all \( x \geq 0 \)\n\n\[ \frac{a}{{\left( x + b\right) }^{2}} + \frac{b}{{\left( x + c\right) }^{2}} + \frac{c}{{\left( x + a\right) }^{2}} \geq \frac{1}{{\left( x + ab + bc + ca\right) }^{2}}\left( \star \right) \]\n\nIndeed, by Cauchy-Schwarz inequality, we obta... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{b\left( {1 + {ab}}\right) } + \frac{1}{c\left( {1 + {bc}}\right) } + \frac{1}{a\left( {1 + {ca}}\right) } \geq \frac{1}{\sqrt[3]{abc}\left( {1 + \sqrt[3]{{a}^{2}{b}^{2}{c}^{2}}}\right) } \]\n\n(1) | SOLUTION. Taking into account example ??, we have\n\n\[ \frac{a}{{\left( x + ab\right) }^{2}} + \frac{b}{{\left( x + bc\right) }^{2}} + \frac{c}{{\left( x + ca\right) }^{2}} \geq \frac{3\sqrt[3]{abc}}{{\left( x + \sqrt[3]{{a}^{2}{b}^{2}{c}^{2}}\right) }^{2}}. \]\n\nIntegrating on \( \left\lbrack {0,1}\right\rbrack \), ... | No |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{4a} + \frac{1}{4b} + \frac{1}{4c} + \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \geq \frac{3}{{3a} + b} + \frac{3}{{3b} + c} + \frac{3}{{3c} + a}. \]\n\n(Gabriel Dospinescu) | SOLUTION. There seems to be no purely algebraic solution to this hard inequality, however, the integral method makes up a very impressive one.\n\nAccording to a well-known inequality (see problem ?? in volume I), we have\n\n\[ {\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{2} \geq 3\left( {{x}^{3}y + {y}^{3}z + {z}^{3}x... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{3a} + \frac{1}{3b} + \frac{1}{3c} + \frac{3}{a + b + c} \geq \frac{1}{{2a} + b} + \frac{1}{{2b} + c} + \frac{1}{{2c} + a} + \frac{1}{{2b} + a} + \frac{1}{{2c} + b} + \frac{1}{{2a} + c}. \] | SOLUTION. Starting from Schur inequality, we have\n\n\[ {x}^{3} + {y}^{3} + {z}^{3} + {3xyz} \geq {xy}\left( {x + y}\right) + {yz}\left( {y + z}\right) + {zx}\left( {z + x}\right) . \]\n\nLetting now \( x = {t}^{a}, y = {t}^{b}, z = {t}^{c} \), the above inequality becomes\n\n\[ \frac{1}{t}\left( {{t}^{3a} + {t}^{3b} +... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{5}{{3a} + b} + \frac{5}{{3b} + c} + \frac{5}{{3c} + a} \geq \frac{9}{a + {3b}} + \frac{9}{b + {3c}} + \frac{9}{c + {3a}}. \]\n\n(Pham Kim Hung) | SOLUTION. Let \( k = \frac{5}{4} \) . We start from the following inequality\n\n\[ {x}^{4} + {y}^{4} + {z}^{4} + k\left( {{x}^{3}y + {y}^{3}z + {z}^{3}x}\right) \geq \left( {1 + \sqrt{2}}\right) \left( {x{y}^{3} + y{z}^{3} + z{x}^{3}}\right) \]\n\n(1)\n\nDenoting \( x = {t}^{a}, y = {t}^{b}, z = {t}^{c} \), we obtain\n... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{8}{a + b + c} \geq \frac{17}{3}\left( {\frac{1}{{2a} + b} + \frac{1}{{2b} + c} + \frac{1}{{2c} + a}}\right) .\n\]\n\n(Pham Kim Hung) | SOLUTION. Let \( k = \frac{8}{9} \), then we need to prove that\n\n\[ \n\frac{1}{3a} + \frac{1}{3b} + \frac{1}{3c} + \frac{3k}{a + b + c} \geq \left( {k + 1}\right) \left( {\frac{1}{{2a} + b} + \frac{1}{{2b} + c} + \frac{1}{{2c} + a}}\right) .\n\]\n\nAccording to example 1.10.1, we have\n\n\[ \n{x}^{3} + {y}^{3} + {z}^... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{3}{a + b + c} \geq \frac{4}{{5a} + b} + \frac{4}{{5b} + c} + \frac{4}{{5c} + a} + \frac{4}{{5b} + a} + \frac{4}{{5c} + b} + \frac{4}{{5a} + c}. \] | SOLUTION. We can construct it from the primary inequality\n\n\[ {a}^{6} + {b}^{6} + {c}^{6} + {a}^{2}{b}^{2}{c}^{2} \geq \frac{2}{3}\left( {{a}^{5}\left( {b + c}\right) + {b}^{5}\left( {c + a}\right) + {c}^{5}\left( {a + b}\right) }\right) . \] | No |
Let \( a, b, c, d \) be positive real numbers with sum 4. Prove that\n\n\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \geq \frac{8}{\left( {a + b}\right) \left( {c + d}\right) } + \frac{8}{\left( {b + c}\right) \left( {d + a}\right) } + \frac{8}{\left( {c + a}\right) \left( {b + d}\right) } - \frac{8}{a + b ... | SOLUTION. It is easy to realize that this result is obtained from Turkevici's inequality\n\n\[ {a}^{4} + {b}^{4} + {c}^{4} + {d}^{4} + {2abcd} \geq {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{d}^{2} + {d}^{2}{a}^{2} + {a}^{2}{c}^{2} + {b}^{2}{d}^{2}. \]\n | No |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{4a} + \frac{1}{4b} + \frac{1}{4c} + \frac{1}{{3a} + b} + \frac{1}{{3b} + c} + \frac{1}{{3c} + a} \geq \frac{2}{a + {3b}} + \frac{2}{b + {3c}} + \frac{2}{c + {3a}}. \] | SOLUTION. We use the following familiar result (shown in the first volume)\n\n\[ {a}^{4} + {b}^{4} + {c}^{4} + {a}^{3}b + {b}^{3}c + {c}^{3}a \geq 2\left( {a{b}^{3} + b{c}^{3} + c{a}^{3}}\right) \]\n\nto deduce the desired result. The equality holds for \( a = b = c \) . \( \nabla \) | No |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\frac{4}{a} + \frac{4}{b} + \frac{4}{c} + \frac{36}{{2a} + b} + \frac{36}{{2b} + c} + \frac{36}{{2c} + a} \geq \frac{45}{a + {2b}} + \frac{45}{b + {2c}} + \frac{45}{c + {2a}} + \frac{9}{a + b + c}.\n\] | SOLUTION. We use the following familiar result\n\n\[ \n{27}\left( {a{b}^{2} + b{c}^{2} + c{a}^{2} + {abc}}\right) \leq 4{\left( a + b + c\right) }^{3}\n\]\n\nto deduce the desired result. Equality holds for only \( a = b = c \) . | No |
Prove that \( {a}^{4} + {b}^{4} + {c}^{4} + {d}^{4} \geq {28abcd} \) for all \( a, b, c, d > 0 \) satisfying\n\n\[ \n{\left( a + b + c + d\right) }^{2} = 3\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) .\n\] | Solution. (for Turkevici’s inequality) Denote \( m = {a}^{2} + {b}^{2}, n = {c}^{2} + {d}^{2} \) and \( x = {ab}, y = {cd} \) . We can rewrite the inequality in the following form\n\n\[ \n{m}^{2} - 2{x}^{2} + {n}^{2} - 2{y}^{2} + {2xy} \geq {x}^{2} + {y}^{2} + \left( {{m}^{2} - {2x}}\right) \left( {{n}^{2} - {2y}}\righ... | Yes |
Let \( a, b, c, d \) be non-negative positive real numbers with sum 1. Prove that\n\n\[ \n{abc} + {bcd} + {cda} + {dab} \leq \frac{1}{27} + \frac{176}{27}{abcd}.\n\] | Solution. In this problem, we fix \( a + b = m \) and \( c + d = n \) . Let \( x = {ab} \) and \( y = {cd} \) then\n\n\[ \n{abc} + {bcd} + {cda} + {dab} - \frac{1}{27} - \frac{176}{27}{abcd} = {my} + {nx} - \frac{1}{27} - \frac{176xy}{27} = f\left( {x, y}\right)\n\]\n\nis a linear (convex) function in both \( x \) and ... | Yes |
Example 1.14.5. Let \( a, b, c, d \) be non-negative real numbers with sum 4. Prove that\n\n\[ \n{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{d}^{2} + {d}^{2}{a}^{2} + {a}^{2}{c}^{2} + {b}^{2}{d}^{2} + {10abcd} \leq {16}.\n\] | Solution. We fix \( a + b = m \) and \( c + d = n \) . Let \( {ab} = x \) and \( {cd} = y \) then\n\n\[ \n\mathop{\sum }\limits_{{cyc}}{a}^{2}{b}^{2} + {10abcd} = {x}^{2} + {y}^{2} + \left( {{m}^{2} - {2x}}\right) \left( {{n}^{2} - {2y}}\right) + {10xy}\n\]\n\nare convex functions in each variable \( x \) and \( y \) .... | Yes |
Let \( a, b, c, d \) be non-negative real numbers with sum 4. Prove that\n\n\[ \left( {1 + {3a}}\right) \left( {1 + {3b}}\right) \left( {1 + {3c}}\right) \left( {1 + {3d}}\right) \leq {125} + {131abcd} \] | Solution. We fix \( a + b = m \) and \( c + d = n \) . Let \( x = {ab} \) and \( y = {cd} \) (we regard \( x \) and \( y \) as variables). The inequality becomes\n\n\[ \left( {1 + {9y} + {3n}}\right) \left( {1 + {9x} + {3m}}\right) - {125} - {131xy} \geq 0. \]\n\nThis expression is a linear (and also convex) function i... | Yes |
Theorem 5 (SIP theorem). Let \( f \) be a twice diffirentiable function on \( \mathbb{R} \) with a single inflection point. For a fixed real number \( S \), we denote\n\n\[ g\left( x\right) = \left( {n - 1}\right) f\left( x\right) + f\left( \frac{S - x}{n - 1}\right) . \]\n\nFor all real numbers \( {x}_{1},{x}_{2},\ldo... | PROOF. First we will prove that\n\n\[ f\left( {x}_{1}\right) + f\left( {x}_{2}\right) + \ldots + f\left( {x}_{n}\right) \geq \mathop{\inf }\limits_{{x \in \mathbb{R}}}g\left( x\right) . \]\n\nAssume that \( a \) is the single inflection point of \( f\left( x\right) \) . Denote \( {\mathbb{I}}_{1} = \lbrack a, + \infty ... | Yes |
Let \( a, b, c, d \) be positive real numbers such that \( {abcd} = 1 \) . Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {d}^{3} + {12} \geq 2\left( {a + b + c + d + {abc} + {bcd} + {cda} + {dab}}\right) .\n\]\n\n(Pham Kim Hung) | Solution. We have to prove that \( f\left( x\right) + f\left( y\right) + f\left( z\right) + f\left( t\right) \geq 4 \) where \( x, y, z, t \) are \( \ln a,\ln b,\ln c,\ln d \) respectively and\n\n\[ \nf\left( x\right) = {e}^{3x} - 2{e}^{x} - 2{e}^{-x}.\n\]\n\nClearly\n\n\[ \n{f}^{\prime \prime }\left( x\right) = 9{e}^{... | Yes |
Let \( a, b, c, d \) be positive real numbers with sum 4. Prove that \[ 9\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}\right) + {56} \geq {15}\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) . \] | Solution. We have to prove that \( f\left( a\right) + f\left( b\right) + f\left( c\right) + f\left( d\right) \geq 0 \) where \[ f\left( x\right) = \frac{9}{x} - {15}{x}^{2}. \] Since \( {f}^{\prime \prime }\left( x\right) = \frac{18}{{x}^{3}} - {30} \) has exactly one positive real root, we infer that \( f\left( x\righ... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers such that \( {a}_{1}{a}_{2}\ldots {a}_{n} = 1 \) . Prove that \[ \frac{1}{n - 1 + {a}_{1}} + \frac{1}{n - 1 + {a}_{2}} + \ldots + \frac{1}{n - 1 + {a}_{n}} \leq 1. \] | SOLUTION. We have to prove that \[ f\left( {x}_{1}\right) + f\left( {x}_{2}\right) + \ldots + f\left( {x}_{n}\right) \leq 1 \] where \( {x}_{i} = \ln {a}_{i}\forall i \in \{ 1,2,\ldots, n\} \) and \[ f\left( x\right) = \frac{1}{n - 1 + {e}^{x}}. \] We have \[ {f}^{\prime \prime }\left( x\right) = \frac{2\left( {{e}^{x}... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers with product 1 . Prove that\n\n\[ \n{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2} - n \geq \frac{2n}{n - 1}\sqrt[n]{n - 1}\left( {{a}_{1} + {a}_{2} + \ldots + {a}_{n} - n}\right) .\n\]\n\n(Gabriel Dospinescu, Calin Popa) | SOLUTION. For \( k = \frac{2n}{n - 1}\sqrt[n]{n - 1} \), we consider the following function\n\n\[ \nf\left( x\right) = {e}^{2x} - k{e}^{x}.\n\]\n\nWe have to prove that \( f\left( {x}_{1}\right) + f\left( {x}_{2}\right) + \ldots + f\left( {x}_{n}\right) \geq \left( {1 - k}\right) n \) where \( {x}_{i} = \ln {a}_{i}\for... | No |
Example 1.14.12. Let \( a, b, c, d \) be positive real numbers with sum 4. Prove that\n\n\[ \left( {1 + {a}^{2}}\right) \left( {1 + {b}^{2}}\right) \left( {1 + {c}^{2}}\right) \left( {1 + {d}^{2}}\right) \geq \frac{{10}^{4}}{{9}^{3}}. \] | SOLUTION. We have to prove that\n\n\[ f\left( a\right) + f\left( b\right) + f\left( c\right) + f\left( d\right) \geq 4\ln {10} - 3\ln 9, \]\n\nwhere \( f\left( x\right) = \ln \left( {1 + {x}^{2}}\right) \) . Since\n\n\[ {f}^{\prime \prime }\left( x\right) = \frac{2\left( {1 - {x}^{2}}\right) }{{\left( 1 + {x}^{2}\right... | Yes |
Lemma 2. Suppose that \( m, n \) are two non-negative real constants, then the system of equations\n\n\[ \left\{ \begin{array}{l} x + y + z = m \\ {xy} + {yz} + {zx} = n \end{array}\right. \]\n\nhas a solution \( \left( {x, y, z}\right) = \left( {a, b, c}\right) \), with \( a, b, c \geq 0 \) if and only if \( {m}^{2} \... | Proof. The second lemma is quite obvious, therefore we will prove the first one.\n\nDenote \( x = a - 1, y = b - 1, z = c - 1 \) then \( x + y + z = r - 1 \) and \( {x}^{2} + {y}^{2} + {z}^{2} = {\left( r - 1\right) }^{2} \), so we infer \( {xy} + {yz} + {zx} = 0 \) . We also have that \( x, y, z \) are the real roots ... | No |
Lemma 3. Suppose that \( a, b, c \) are three non-negative real numbers satisfying \( a + b + c = m, ab + bc + ca = n \), where \( m \) and \( n \) are two non-negative real constants. If \( m^2 \geq 4n \) then the minimum value of abc is 0, with equality for \( a = 0 \) and \( \left( b, c \right) = \left( \frac{m + \s... | Proof. We consider the following cases\n\n(i) The first case. If \( m^2 \leq 4n \) then there exist two numbers \( a_0, b_0 \) such that \( a + b = m, ab = 4n \), therefore \( \left( a, b, c \right) = \left( a_0, b_0, 0 \right) \) satisfies the system\n\n\[ \left\{ \begin{array}{l} a + b + c = m \\ ab + bc + ca = n \en... | Yes |
Lemma 4. Suppose that \( a, b, c \) are three non-negative real numbers satisfying \( a + \) \( b + c = m,{ab} + {bc} + {ca} = n \), where \( m \) and \( n \) are two non-negative real constants and \( {m}^{2} \geq {3n} \) . The maximum value of \( {abc} \) is attained for \( \left( {a, b, c}\right) = \) | \[ \left( {\frac{m + 2\sqrt{{m}^{2} - {3n}}}{3},\frac{m - \sqrt{{m}^{2} - {3n}}}{3},\frac{m - \sqrt{{m}^{2} - {3n}}}{3}}\right) \text{up to permutation.} \] | Yes |
Corollary 2. Suppose that \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) are non-negative real numbers satisfying\n\n\[ \n{x}_{1} + {x}_{2} + \ldots + {x}_{n} = \text{ const },{x}_{1}^{2} + {x}_{2}^{2} + \ldots + {x}_{n}^{2} = \text{ const }\n\]\n\nand \( f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) a continuous, symmetr... | Proof. To prove the above corollaries, we only show the hardest, that is the second part of the second corollary (and other parts are proved similarly).\n\n- Suppose that \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) are non-negative real numbers satisfying\n\n\[ \n{x}_{1} + {x}_{2} + \ldots + {x}_{n} = \text{ const },{x}_{1}^... | Yes |
Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} + {d}^{4} + {2abcd} \geq {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{d}^{2} + {d}^{2}{a}^{2} + {a}^{2}{c}^{2} + {b}^{2}{d}^{2}.\n\]\n\n(Turkevici's inequality) | Solution. If we fix \( a + b + c = \) const and \( {a}^{2} + {b}^{2} + {c}^{2} = \) const, then\n\n\[ \n\text{RHS} - \text{LHS} = {\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2} - 3{\left( ab + bc + ca\right) }^{2} + {6abc}\left( {a + b + c}\right) + {2abcd} - {d}^{2}\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right)\n\]\n\n... | Yes |
Let \( a, b, c, d \) be non-negative real numbers with sum 4. Prove that\n\n\[ \left( {1 + {2a}}\right) \left( {1 + {2b}}\right) \left( {1 + {2c}}\right) \left( {1 + {2d}}\right) \leq {10}\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) + {41abcd}. \] | Solution. First, notice that if \( a \leq b \leq c \leq d \) and \( c \leq 1/3 \) then \( a, b \leq 1/3 \) and \( d \geq 3 \) and we are done because\n\n\[ {10}\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) + {41abcd} \geq {90} > \left( {1 + {2a}}\right) \left( {1 + {2b}}\right) \left( {1 + {2c}}\right) \left( {... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be non-negative real numbers such that \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = n \) . Prove that\n\n\[ \n{\left( {x}_{1}{x}_{2}\ldots {x}_{n}\right) }^{\frac{1}{\sqrt{n - 1}}}\left( {{x}_{1}^{2} + {x}_{2}^{2} + \ldots + {x}_{n}^{2}}\right) \leq n.\n\]\n\n(Vasile Cirtoaje) | SOLUTION. If we fix \( {x}_{1} + {x}_{2} + {x}_{3} \) and \( {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} \) then the left-hand expression of the above inequality is clearly a strictly increasing function of \( {x}_{1}{x}_{2}{x}_{3} \), so, according to the corollary 2 of \( n \) SMV theorem, we conclude that it suffices to... | Yes |
Let \( a, b, c, d \) be positive real numbers such that that\n\n\[ 2{\left( a + b + c + d\right) }^{2} = 5\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) .\n\]\nFind the minimum value of\n\n\[ P = \frac{{a}^{4} + {b}^{4} + {c}^{4} + {d}^{4}}{abcd}. \] | SOLUTION. First we guess that the equality holds for \( b = c = d \) (in this case, we find out \( a = 2 + \sqrt{5}, b = c = d = 1 \) and permutations) and this is the key to the solution. Indeed, denote \( k = {\left( 2 + \sqrt{5}\right) }^{3} + {\left( 2 + \sqrt{5}\right) }^{-1} > {78} \), we will prove\n\n\[ {a}^{4}... | Yes |
Suppose that \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are positive real numbers satisfying\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{n} = {a}_{1}{}^{-1} + {a}_{2}{}^{-1} + \ldots + {a}_{n}{}^{-1} = n + 2.\n\]\n\nFind the minimum and maximum value of\n\n\[ \nP = {a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2} + 2{a}_{1}{a... | Solution. Without loss of generality, we may assume that \( {a}_{1} \geq {a}_{2} \geq \ldots \geq {a}_{n - 1} \geq \) \( {a}_{n} \) . We fix \( s = {a}_{1} + {a}_{2} + {a}_{n} \) and \( r = \frac{1}{{a}_{1}} + \frac{1}{{a}_{2}} + \frac{1}{{a}_{n}} \) . Denote \( x = {a}_{1}{a}_{2}{a}_{n} \) and \( p = \) \( {a}_{3}{a}_... | Yes |
Let \( a, b, c, d \) be non-negative real numbers with sum 1. Prove that\n\n\[\n2\left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) \geq {27}\sqrt[3]{\left( {{a}^{2} + {b}^{2}}\right) \left( {{a}^{2} + {c}^{2}}\right) \left( {{a}^{2} + {d}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) \left( {{b}^{2} + {d}^{2}}\righ... | Solution. We will prove a homogeneous inequality as follows\n\n\[\n2{\left( \mathop{\sum }\limits_{{cyc}}a\right) }^{2}\left( {\mathop{\sum }\limits_{{cyc}}{a}^{2}}\right) \geq {27}\sqrt[3]{\mathop{\prod }\limits_{\text{sym }}\left( {{a}^{2} + {b}^{2}}\right) }.\n\]\n\nWLOG, assume that \( a \geq b \geq c \geq d \) . I... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive real numbers such that \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = n \) . Prove that \[ \frac{1}{{x}_{1}} + \frac{1}{{x}_{2}} + \ldots + \frac{1}{{x}_{n}} + \frac{{2n}\sqrt{n - 1}}{{x}_{1}^{2} + {x}_{2}^{2} + \ldots + {x}_{n}^{2}} \geq n + 2\sqrt{n - 1}. \] | Solution. We fix the sums \( {x}_{1} + {x}_{2} + {x}_{3},{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} \) and fix the \( n - 3 \) numbers \( {x}_{3},{x}_{4},\ldots ,{x}_{n} \) . Clearly, \( {x}_{1}{x}_{2} + {x}_{2}{x}_{3} + {x}_{3}{x}_{1} \) is a constant and \[ \frac{1}{{x}_{1}} + \frac{1}{{x}_{2}} + \frac{1}{{x}_{3}} = \fr... | No |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n}\left( {n \geq 4}\right) \) be positive real numbers such that \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = n \) . Prove that \[ \frac{1}{{x}_{1}} + \frac{1}{{x}_{2}} + \ldots + \frac{1}{{x}_{n}} + \frac{\sqrt{n}\left( {\sqrt{n + 4} + 2\sqrt{n - 1}}\right) }{\sqrt{{x}_{1}^{2} + {x}_{2... | SOLUTION. Similarly with the previous example, if we fix \( {x}_{1} + {x}_{2} + {x}_{3},{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} \) and fix \( {x}_{4},{x}_{4},\ldots ,{x}_{n} \) then the left hand side of the inequality is a decreasing function of \( {x}_{1}{x}_{2}{x}_{3} \) . So we may assume that \( {x}_{1} = {x}_{2} ... | Yes |
Proposition 3. Let \( \left( a\right) = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \left( b\right) = \left( {{b}_{2},{b}_{2},\ldots ,{b}_{n}}\right) \) be two sequences of real numbers. We have that \( \left( {a}^{ * }\right) \) majorizes \( \left( b\right) \) if the following conditions are fulfilled\n\... | These properties are quite obvious: they can be proved directly from the definition of Majorization. | No |
Proposition 4. If \( {x}_{1} \geq {x}_{2} \geq \ldots \geq {x}_{n} \) and \( {y}_{1} \geq {y}_{2} \geq \ldots \geq {y}_{n} \) are positive real numbers such that \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = {y}_{1} + {y}_{2} + \ldots + {y}_{n} \) and \( \frac{{x}_{i}}{{x}_{j}} \geq \frac{{y}_{i}}{{y}_{j}}\forall i < j \),... | Proof. To prove this assertion, we will use induction. Because \( \frac{{x}_{i}}{{x}_{1}} \leq \frac{{y}_{i}}{{y}_{1}} \) for all \( i \in \) \( \{ 1,2,\ldots, n\} \), we get that\n\n\[ \frac{{x}_{1} + {x}_{2} + \ldots + {x}_{n}}{{x}_{1}} \leq \frac{{y}_{1} + {y}_{2} + \ldots + {y}_{n}}{{y}_{1}} \Rightarrow {x}_{1} \ge... | No |
Theorem 7 (Symmetric Majorization Criterion). Suppose that \( \left( a\right) = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \left( b\right) = \left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) \) are two sequences of real numbers; then \( \left( {a}^{ * }\right) \gg \left( {b}^{ * }\right) \) if and only if ... | Proof. To prove this theorem, we need to prove the following.\n\n(i). Necessary condition. Suppose that \( \left( {a}^{ * }\right) \gg \left( {b}^{ * }\right) \), then we need to prove that for all real numbers \( x \)\n\n\[ \left| {{a}_{1} - x}\right| + \left| {{a}_{2} - x}\right| + \ldots + \left| {{a}_{n} - x}\right... | Yes |
Theorem 8 (Karamata inequality). If (a) and (b) two numbers sequences for which \( \left( {a}^{ * }\right) \gg \left( {b}^{ * }\right) \) and \( f \) is a convex function twice differentiable on \( \mathbb{I} \) then\n\n\[ f\left( {a}_{1}\right) + f\left( {a}_{2}\right) + \ldots + f\left( {a}_{n}\right) \geq f\left( {b... | Proof. WLOG, assume that \( {a}_{1} \geq {a}_{2} \geq \ldots \geq {a}_{n} \) and \( {b}_{1} \geq {b}_{2} \geq \ldots \geq {b}_{n} \) . The inductive hypothesis yields \( \left( a\right) = \left( {a}^{ * }\right) \gg \left( {b}^{ * }\right) = \left( b\right) \) . Notice that \( f \) is a twice differentiable function on... | Yes |
If \( f \) is a convex function then\n\n\[ f\left( a\right) + f\left( b\right) + f\left( c\right) + f\left( \frac{a + b + c}{3}\right) \geq \frac{4}{3}\left( {f\left( \frac{a + b}{2}\right) + f\left( \frac{b + c}{2}\right) + f\left( \frac{c + a}{2}\right) }\right) .\n\]\n(Popoviciu's inequality) | Solution. WLOG, suppose that \( a \geq b \geq c \) . Consider the following number sequences\n\n\[ \left( x\right) = \left( {a, a, a, b, t, t, t, b, b, c, c, c}\right) \;;\;\left( y\right) = \left( {\alpha ,\alpha ,\alpha ,\alpha ,\beta ,\beta ,\beta ,\beta ,\gamma ,\gamma ,\gamma ,\gamma }\right) \;;\n\]\n\nwhere\n\n\... | Yes |
Example 1.17.2 (Jensen Inequality). If \( f \) is a convex function then\n\n\[ f\left( {a}_{1}\right) + f\left( {a}_{2}\right) + \ldots + f\left( {a}_{n}\right) \geq {nf}\left( \frac{{a}_{1} + {a}_{2} + \ldots + {a}_{n}}{n}\right) . \] | SOLUTION. We use property 1 of majorization. Suppose that \( {a}_{1} \geq {a}_{2} \geq \ldots \geq {a}_{n} \), then we have \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \gg \left( {a, a,\ldots, a}\right) \) with \( a = \frac{1}{n}\left( {{a}_{1} + {a}_{2} + \ldots + {a}_{n}}\right) \) . Our problem is directly de... | Yes |
Let \( a, b, c, x, y, z \) be six real numbers in \( \mathbb{I} \) satisfying\n\n\[ \na + b + c = x + y + z,\max \left( {a, b, c}\right) \geq \max \left( {x, y, z}\right) ,\min \left( {a, b, c}\right) \leq \min \left( {x, y, z}\right) ,\n\]\n\nthen for every convex function \( f \) on \( \mathbb{I} \), we have\n\n\[ \n... | Solution. Assume that \( x \geq y \geq z \) . The assumption implies \( {\left( a, b, c\right) }^{ * } \gg \left( {x, y, z}\right) \) and the conclusion follows from Karamata inequality. \( \nabla \) | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers. Prove that\n\n\[ \left( {1 + {a}_{1}}\right) \left( {1 + {a}_{2}}\right) \ldots \left( {1 + {a}_{n}}\right) \leq \left( {1 + \frac{{a}_{1}^{2}}{{a}_{2}}}\right) \left( {1 + \frac{{a}_{2}^{2}}{{a}_{3}}}\right) \ldots \left( {1 + \frac{{a}_{n}^{2}}{{a}_{... | SOLUTION. Our inequality is equivalent to\n\n\[ \ln \left( {1 + {a}_{1}}\right) + \ln \left( {1 + {a}_{2}}\right) + \ldots + \ln \left( {1 + {a}_{n}}\right) \leq \ln \left( {1 + \frac{{a}_{1}^{2}}{{a}_{2}}}\right) + \ln \left( {1 + \frac{{a}_{2}^{2}}{{a}_{3}}}\right) + \ldots + \ln \left( {1 + \frac{{a}_{n}^{2}}{{a}_{1... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers. Prove that\n\n\[ \frac{{a}_{1}^{2}}{{a}_{2}^{2} + \ldots + {a}_{n}^{2}} + \ldots + \frac{{a}_{n}^{2}}{{a}_{1}^{2} + \ldots + {a}_{n - 1}^{2}} \geq \frac{{a}_{1}}{{a}_{2} + \ldots + {a}_{n}} + \ldots + \frac{{a}_{n}}{{a}_{1} + \ldots + {a}_{n - 1}}. \] | Solution. For each \( i \in \{ 1,2,\ldots, n\} \), we denote\n\n\[ {y}_{i} = \frac{{a}_{i}}{{a}_{1} + {a}_{2} + \ldots + {a}_{n}},{x}_{i} = \frac{{a}_{i}^{2}}{{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2}} \]\n\nthen \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = {y}_{1} + {y}_{2} + \ldots + {y}_{n} = 1 \) . We need to p... | Yes |
Suppose that \( \\left( {{a}_{1},{a}_{2},\\ldots ,{a}_{2n}}\\right) \) is a permutation of \( \\left( {{b}_{1},{b}_{2},\\ldots ,{b}_{2n}}\\right) \) which satisfies \( {b}_{1} \\geq {b}_{2} \\geq \\ldots \\geq {b}_{2n} \\geq 0 \) . Prove that\n\n\[ \n\\left( {1 + {a}_{1}{a}_{2}}\\right) \\left( {1 + {a}_{3}{a}_{4}}\\ri... | Solution. Denote \( f\\left( x\\right) = \\ln \\left( {1 + {e}^{x}}\\right) \) and \( {x}_{i} = \\ln {a}_{i},{y}_{i} = \\ln {b}_{i} \) . We need to prove that\n\n\[ \nf\\left( {{x}_{1} + {x}_{2}}\\right) + f\\left( {{x}_{3} + {x}_{4}}\\right) + \\ldots + f\\left( {{x}_{{2n} - 1} + {x}_{2n}}\\right) \n\]\n\n\[ \n\\leq f... | Yes |
Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} + {d}^{4} + {2abcd} \geq {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{d}^{2} + {d}^{2}{a}^{2} + {a}^{2}{c}^{2} + {b}^{2}{d}^{2}. \n\]\n\n(Turkevici's inequality) | To prove this problem, we use the following lemma\n\n* For all real numbers \( x, y, z, t \) then\n\n\[ \n2\left( {\left| x\right| + \left| y\right| + \left| z\right| + \left| t\right| }\right) + \left| {x + y + z + t}\right| \geq \left| {x + y}\right| + \left| {y + z}\right| + \left| {z + t}\right| + \left| {t + x}\ri... | No |
Example 1. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{a + c}{a + b} + \frac{b + c}{b + a} + \frac{c + a}{c + b}. \]\n\nMathlinks Contests | Solution. Without loss of generality, assume that \( c = \min \left( {a, b, c}\right) \) . Note that for \( x, y, z > 0 \) we have\n\n\[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} - 3 = \frac{1}{xy}{\left( x - y\right) }^{2} + \frac{1}{xz}\left( {x - z}\right) \left( {y - z}\right) .\n\nTherefore the given inequality can ... | Yes |
Let \( a, b, c \) be positive real numbers such that \( a \geq b \geq c \) . Prove that\n\n\[ \n{a}^{2}b\left( {a - b}\right) + {b}^{2}c\left( {b - c}\right) + {c}^{2}a\left( {c - a}\right) \geq 0.\n\] | Solution. We have\n\n\[ \nf\left( {a, b, c}\right) = {a}^{2}b\left( {a - b}\right) + {b}^{2}c\left( {b - c}\right) + {c}^{2}a\left( {c - a}\right) \n\]\n\n\[ \n= \left\lbrack {{a}^{2}b\left( {a - b}\right) - a{b}^{2}\left( {a - b}\right) }\right\rbrack + \left\lbrack {{b}^{2}c\left( {b - c}\right) - a{b}^{2}\left( {b -... | Yes |
Example 3. Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\frac{a + b}{b + c} + \frac{b + c}{c + a} + \frac{c + a}{a + b} + \frac{3\left( {{ab} + {bc} + {ca}}\right) }{{\left( a + b + c\right) }^{2}} \geq 4.\n\] | Solution. Without loss of generality, assume that \( c = \min \left( {a, b, c}\right) \) . We have\n\n\[ \n\frac{a + b}{b + c} + \frac{b + c}{c + a} + \frac{c + a}{a + b} - 3 = \frac{1}{\left( {a + c}\right) \left( {b + c}\right) }{\left( a - b\right) }^{2} + \frac{1}{\left( {a + b}\right) \left( {b + c}\right) }\left(... | Yes |
Problem 1 . Let a, b, c, x, y, z be six real (not necessarily nonnegative) numbers. Assume that \( a \geq b \geq c \) . Also, assume that either \( x \geq y \geq z \) or \( x \leq y \leq z \) . Then,\n\n\[ \left( {\mathop{\sum }\limits_{\text{cyc }}\left( {a - b}\right) \left( {a - c}\right) }\right) \cdot \left( {\mat... | Solution:\n\nThe inequality we have to prove rewrites as\n\n\[ \left( {\mathop{\sum }\limits_{\text{cyc }}\left( {a - b}\right) \left( {a - c}\right) }\right) \cdot \left( {\mathop{\sum }\limits_{\text{cyc }}{x}^{2}\left( {a - b}\right) \left( {a - c}\right) }\right) - {\left( \mathop{\sum }\limits_{\text{cyc }}x\left(... | Yes |
Problem 2(Darij Grinberg). If \( p \) is an even nonnegative integer, then the inequality \( \mathop{\sum }\limits_{\text{cyc }}{a}^{p}\left( {a - b}\right) \left( {a - c}\right) \geq 0 \) holds for arbitary reals \( a, b, c \) . | Solution: Since the inequality in question is symmetric, we can WLOG assume that \( a \geq b \geq c \) . Since \( \mathrm{p} \) is an even nonnegative integer, we have \( p = {2n} \) for some nonnegative integer \( n \) .\n\n\[ \n\text{Define a function sign by sign}t = \left\{ \begin{matrix} - 1, & \text{ if }t < 0 \\... | Yes |
If \( a, b, c \) are nonnegative real numbers, then \[ {a}^{4} + {b}^{4} + {c}^{4} + {abc}\left( {a + b + c}\right) \geq \sum {bc}\left( {b + c}\right) \sqrt{{b}^{2} - {bc} + {c}^{2}}. \] | Firstly Solution: First, we have: \( \left( {b + c}\right) \sqrt{{b}^{2} - {bc} + {c}^{2}} \geq {b}^{2} + {c}^{2} \Leftrightarrow {bc}{\left( b - c\right) }^{2} \geq 0 \) and then, \[ \left( {b + c}\right) \sqrt{{b}^{2} - {bc} + {c}^{2}} - \left( {{b}^{2} + {c}^{2}}\right) = \frac{{bc}{\left( b - c\right) }^{2}}{\left(... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{\left( {b + c - a}\right) \left( {c + a - b}\right) \left( {a + b - c}\right) }{2abc} \leq 2. \] | We have\n\n\[ 1 - \frac{\left( {b + c - a}\right) \left( {c + a - b}\right) \left( {a + b - c}\right) }{abc} = \sum \frac{a\left( {a - b}\right) \left( {a - c}\right) }{abc} = \sum \frac{\left( {a - b}\right) \left( {a - c}\right) }{bc}. \]\n\nand\n\n\[ \sum \frac{2a}{b + c} - 3 = \sum \frac{\left( {a - b}\right) \left... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{3}}{2{a}^{2} + {b}^{2}} + \frac{{b}^{3}}{2{b}^{2} + {c}^{2}} + \frac{{c}^{3}}{2{c}^{2} + {a}^{2}} \geq \frac{a + b + c}{3}. \]\n | Solution:\n\nWe have that\n\n\[ \sum \frac{{a}^{3} - a{b}^{2}}{2{a}^{2} + {b}^{2}} \geq 0 \Leftrightarrow \sum \left( {{a}^{3} - a{b}^{2}}\right) \left( {2{b}^{2} + {c}^{2}}\right) \left( {2{c}^{2} + {a}^{2}}\right) \geq 0 \]\n\n\[ \Leftrightarrow 3\sum {a}^{3}{b}^{2}{c}^{2} + 2\sum {a}^{3}{c}^{4} + 2\sum {a}^{5}{b}^{2... | Yes |
Problem 10(Nguyen Duy Tung). Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \n\frac{4a}{a + b} + \frac{4b}{b + c} + \frac{4c}{c + a} + \frac{a{b}^{2} + b{c}^{2} + c{a}^{2} + {abc}}{{a}^{2}b + {b}^{2}c + {c}^{2}a + {abc}} \geq 7 \n\] | Solution:\n\nWe have that\n\n\[ \n2\left( {3 - \frac{\left( {a - b}\right) \left( {b - c}\right) \left( {c - a}\right) }{\left( {a + b}\right) \left( {b + c}\right) \left( {c + a}\right) }}\right) + \left( {\frac{a{b}^{2} + b{c}^{2} + c{a}^{2} + {abc}}{{a}^{2}b + {b}^{2}c + {c}^{2}a + {abc}} - 1}\right) \geq 6 \n\]\n\n... | Yes |
Problem 11(Vasile Cirtoaje). Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + 2\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) \geq 3\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) \n\] | ## Solution:\n\nWLOG, we may assume \( b \) is number betwen to two numbers \( a \) and \( c \) .\n\nIf \( a \geq b \geq c \) then \( 2\left( {{a}^{2}b + {b}^{2}c + {c}^{2}a}\right) \geq 2\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) \)\n\nAnd \( {a}^{3} + {b}^{3} + {c}^{3} \geq a{b}^{2} + b{c}^{2} + c{a}^{2} \) so th... | Yes |
Problem 15(Nguyen Duy Tung). Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{3abc}{2\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) } \geq 2. \] | With three variable inequality we have different nice method is SOS-Schur (SS) based on that equality\n\n\[ {a}^{2} + {b}^{2} + {c}^{2} - {ab} - {bc} - {ca} = \left( {{a}^{2} + {b}^{2} - {2ab}}\right) + \left( {{c}^{2} - {ca} - {cb} + {ab}}\right) = {\left( a - b\right) }^{2} + \left( {a - c}\right) \left( {b - c}\righ... | No |
Problem 16(Nguyen Duy Tung). Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \frac{{a}^{2} + {bc}}{{b}^{2} + {c}^{2}} + \frac{{b}^{2} + {ca}}{{c}^{2} + {a}^{2}} + \frac{{c}^{2} + {ab}}{{a}^{2} + {b}^{2}} \geq \frac{5}{2} + \frac{4{a}^{2}{b}^{2}{c}^{2}}{\left( {{a}^{2} + {b}^{2}}\right) \left( {{b}^{2} +... | ## Solution:\n\nThe inequality equivalent to\n\n\[ 2\sum \left\lbrack {\left( {{a}^{2} + {bc}}\right) \left( {{a}^{2} + {b}^{2}}\right) \left( {{a}^{2} + {c}^{2}}\right) }\right\rbrack \geq \left( {{a}^{2} + {b}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) \left( {{c}^{2} + {a}^{2}}\right) + 8{a}^{2}{b}^{2}{c}^{2}. \]... | Yes |
Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \frac{{a}^{2} + {bc}}{{b}^{2} + {c}^{2}} + \frac{{b}^{2} + {ca}}{{c}^{2} + {a}^{2}} + \frac{{c}^{2} + {ab}}{{a}^{2} + {b}^{2}} \geq \frac{5}{2} + \frac{4{a}^{2}{b}^{2}{c}^{2}}{\left( {{a}^{2} + {b}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) \left( {{c}^... | The inequality equivalent to\n\n\[ 2\sum \left\lbrack {\left( {{a}^{2} + {bc}}\right) \left( {{a}^{2} + {b}^{2}}\right) \left( {{a}^{2} + {c}^{2}}\right) }\right\rbrack \geq \left( {{a}^{2} + {b}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) \left( {{c}^{2} + {a}^{2}}\right) + 8{a}^{2}{b}^{2}{c}^{2}. \]\n\n\[ \Leftrigh... | Yes |
Problem 10. Let \( a, b \) and \( c \) are positive numbers. Prove that\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3\sqrt{\frac{{a}^{2} + {b}^{2} + {c}^{2}}{{ab} + {ac} + {bc}}} \] | Solution:\n\nNotice that if \( a \geq b \geq c \) then\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\nso it enough to consider the case \( a \geq b \geq... | Yes |
Problem 11(Nguyen Duy Tung). Let \( a, b, c \) be positive real numebrs . Prove that\n\n\[ \frac{a\left( {b + c}\right) }{{b}^{2} + {c}^{2}} + \frac{b\left( {c + a}\right) }{{c}^{2} + {a}^{2}} + \frac{c\left( {a + b}\right) }{{a}^{2} + {b}^{2}} \geq 2 + \frac{8{a}^{2}{b}^{2}{c}^{2}}{\left( {{a}^{2} + {b}^{2}}\right) \l... | Solution:\n\nThe inequality equivalent to\n\n\[ \sum \left\lbrack {a\left( {b + c}\right) \left( {{a}^{2} + {b}^{2}}\right) \left( {{a}^{2} + {c}^{2}}\right) }\right\rbrack \geq 2\left( {{a}^{2} + {b}^{2}}\right) \left( {{b}^{2} + {c}^{2}}\right) \left( {{c}^{2} + {a}^{2}}\right) + 8{a}^{2}{b}^{2}{c}^{2}. \]\n\n\[ \Lef... | Yes |
Problem 12(Vo Quoc Ba Can). Let \( a, b, c \) be positive real numbers . Prove that\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{8}{3} \cdot \frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} \geq \frac{17}{3}. \] | ## Solution:\n\nThe inequality equivalent\n\n\[ \left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3}\right) + \frac{8}{3} \cdot \left( {\frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} - 1}\right) \geq 0. \]\n\n\[ \Leftrightarrow M{\left( a - b\right) }^{2} + N\left( {a - b}\right) \left( {a - c}\right) \geq 0. \... | Yes |
Problem 14(Nguyen Duy Tung). Let \( a, b, c \) be three side-lengths of a triangle. Prove that \[ 2\left( {\frac{{a}^{2}}{b} + \frac{{b}^{2}}{c} + \frac{{c}^{2}}{a}}\right) \geq a + b + c + \frac{{b}^{2}}{a} + \frac{{c}^{2}}{b} + \frac{{a}^{2}}{c}. \] | Solution:\n\nClearly, this one is equivalent to\n\n\[ \frac{{\left( a - b\right) }^{2}}{b} + \frac{{\left( b - c\right) }^{2}}{c} + \frac{{\left( c - a\right) }^{2}}{a} \geq \frac{\left( {a - b}\right) \left( {b - c}\right) \left( {c - a}\right) \left( {a + b + c}\right) }{abc}. \]\n\n\[ \Leftrightarrow \mathop{\sum }\... | Yes |
Problem 15(Vo Quoc Ba Can). Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{{21}\left( {{ab} + {bc} + {ca}}\right) }{{\left( a + b + c\right) }^{2}} \geq {10}.\n\] | In SS method we have all symmetry and cyclic inequality we sure changes to\n\n\[ \nM{\left( a - b\right) }^{2} + N\left( {a - c}\right) {\left( b - c\right) }^{2} \geq 0 \Leftrightarrow M{\left( a - b\right) }^{2} + N\left( {a - b + b - c}\right) \left( {b - c}\right) \geq 0.\n\]\n\n\[ \n\Leftrightarrow K = M{\left( a ... | No |
Problem 19 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand to that inequality always true\n\n\[ \left( {a + b + c}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\right) + k \cdot \frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} \geq 9 + k. \] | The answre is \( k = 4\sqrt{2} \). | Yes |
Let \( a, b, c \) be nonnegative real numbers. Find the beter constand to that inequality always true\n\n\[ \frac{bc}{{b}^{2} + {c}^{2} + k{a}^{2}} + \frac{ca}{{c}^{2} + {a}^{2} + k{b}^{2}} + \frac{ab}{{a}^{2} + {b}^{2} + k{c}^{2}} \leq \frac{3}{5}. \] | The answre is \( k = 3 \) . | No |
Problem 21 Let \( a, b, c \) be positive real numbers such that \( a + b + c = 3 \) .Prove that\n\n\[ {a}^{2}b + {b}^{2}c + {c}^{2}a \leq 4 \] | Solution: The inequality equivalent to\n\n\[ 2\mathop{\sum }\limits_{{cyc}}{a}^{2}b \leq 8 \]\n\n\[ \Leftrightarrow \left( {\mathop{\sum }\limits_{{cyc}}{a}^{2}b + \mathop{\sum }\limits_{{cyc}}a{b}^{2}}\right) + \left( {\mathop{\sum }\limits_{{cyc}}{a}^{2}b - \mathop{\sum }\limits_{{cyc}}a{b}^{2}}\right) \leq 8 \]\n\n\... | Yes |
Problem 22 Let \( a, b, c \) be positive real numbers such that \( a + b + c = 3 \) .Prove that\n\n\[ {a}^{2}b + {b}^{2}c + {c}^{2}a + 2\left( {a{b}^{2} + b{c}^{2} + c{a}^{2}}\right) \leq 6\sqrt{3} \] | Solution: The inequality equivalent to\n\n\[ \Leftrightarrow 2\mathop{\sum }\limits_{{cyc}}{a}^{2}b + 4\mathop{\sum }\limits_{{cyc}}a{b}^{2} \leq {12}\sqrt{3}. \]\n\n\[ \Leftrightarrow 3\mathop{\sum }\limits_{{sym}}{a}^{2}\left( {b + c}\right) + \left( {\mathop{\sum }\limits_{{cyc}}a{b}^{2} - \mathop{\sum }\limits_{{cy... | Yes |
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