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Proposition 1.2. Assume that all the eigenvalues of the system are genuinely nonlinear (that is, \( {\partial }_{j}{\lambda }_{j}\left( 0\right) \neq 0 \) ). Then the approximate solution \( {\widetilde{u}}^{s} \) is defined for \( 0 \leq t < {\widetilde{T}}_{\varepsilon }^{s} \), with\n\n\[ \varepsilon {\widetilde{T}}... | Proof of Proposition 1.2. For each equation (1.9), the results from Chapter IV (1. a) show that the solution \( {w}_{j} \) has a lifespan of inverse\n\n\[ \max - {\partial }_{j}{\lambda }_{j}\left( 0\right) {\partial }_{\sigma }{w}_{j}\left( {\sigma ,0}\right) + O\left( \varepsilon \right) \]\n\nMoreover,\n\n\[ {w}_{j}... | Yes |
The solution \( v \) of the wave equation with smooth initial data \( f, g \) supported in \( \left| x\right| \leq M \) can be written, for \( \left| x\right| \geq {2M} \), in the form\n\n(2.4)\n\n\[ u\left( {x, t}\right) = {r}^{-\frac{1}{2}\left( {n - 1}\right) }F\left( {r - t,\omega ,{r}^{-1}}\right) . \] | Here \( r = \left| x\right|, x = {r\omega } \) and \( F \) is a smooth function satisfying\n\n(2.5)\n\n\[ \left| {{\partial }_{\rho }^{\alpha }{\partial }_{\omega }^{\beta }{\partial }_{z}^{\gamma }F\left( {\rho ,\omega, z}\right) }\right| \leq {C}_{\alpha \beta \gamma }{\left( 1 + \left| \rho \right| \right) }^{-\frac... | Yes |
Lemma 2.2. Define, with \( {\omega }_{0} = - 1 \) ,\n\n\[ g\left( \omega \right) = \mathop{\sum }\limits_{{0 \leq i, j, k \leq n}}{g}_{ij}^{k}{\omega }_{i}{\omega }_{j}{\omega }_{k} \]\n\nand assume\n\n\[ g\left( \omega \right) ≢ 0. \]\n\nThen for \( \left| x\right| \geq t + M - C \) and \( n = 2 \) ,\n\n\[ {u}^{\left(... | Sketch of the proof of Lemma 2.2.\n\na. From the asymptotic properties of \( {u}^{\left( 1\right) } \) described in Lemma 2.1, we obtain\n\n\[ {Q}^{\left( 2\right) }\left( {x, t}\right) = - g\left( \omega \right) {r}^{-\left( {n - 1}\right) }\left( {{\partial }_{\rho }{F}_{0}{\partial }_{\rho }^{2}{F}_{0}}\right) \left... | Yes |
Lemma 2.3. Inserting (2.16) into (2.1) gives\n\n(2.17)\n\n\[ \sum {g}_{ij}\left( {\nabla u}\right) {\partial }_{ij}^{2}u = \frac{{\varepsilon }^{2}}{{\left( rt\right) }^{\frac{1}{2}}}\left\{ {\frac{}{} - {\partial }_{\rho \tau }^{2}w + g\left( \omega \right) \left( {{\partial }_{\rho }w}\right) \left( {{\partial }_{\rh... | where \( R\left( w\right) \) is a quadratic expression, with smooth coefficients bounded for \( 0 < {\tau }_{0} \leq \tau \leq {\tau }_{1} \), of derivatives \( {\partial }_{\rho ,\omega ,\tau }^{\alpha }w\left( {\left| \alpha \right| \leq 2}\right) \). | No |
Proposition 3.1. For all \( C > 0, N,{N}^{\prime } \in \mathbb{N} \), there exist functions\n\n\[ \n{L}_{q}^{\left( p\right) }\left( {\rho ,\omega, z}\right) ,{R}_{q,{q}^{\prime }}^{\left( p\right) }\left( {\rho ,\omega, z}\right) ,\rho = r - t, z = {r}^{-1} \]\n\n such that the term \( {u}^{\left( p\right) } \) can be... | For instance,\n\n\[ \n{u}^{\left( 1\right) }\left( {x, t}\right) = {r}^{-\frac{1}{2}}{L}_{0}^{\left( 1\right) }\left( {r - t,\omega, z}\right) + {r}^{\left( 1\right) }, \]\n\nwhich is just (2.4); also\n\n\[ \n{u}^{\left( 2\right) }\left( {x, t}\right) = {r}^{-\frac{1}{2}}\left\{ {{L}_{0}^{\left( 2\right) }\left( {r - t... | Yes |
Theorem 3.4. Assume that the function \( - g\left( \omega \right) {\partial }_{\rho }^{2}{F}_{0}\left( {\rho ,\omega }\right) \) has a unique negative minimum at \( \left( {{\rho }_{0},{\omega }_{0}}\right) \), with positive definite Hessian. Then there exists a function \( {\bar{T}}_{\varepsilon }^{a} \) with the two ... | We have in particular \n\n\[ \n{\bar{\tau }}^{a} = \bar{\tau } + {A}_{1}\varepsilon + O\left( {{\varepsilon }^{2}\ell {n\varepsilon }}\right) . \n\] \nThe second constant is \n\n\[ \n{A}_{1} = - {\bar{\tau }}^{2}g\left( {\omega }_{0}\right) {\partial }_{\rho }^{2}{L}_{0}^{\left( 2\right) }\left( {{\rho }_{0},{\omega }_... | No |
Corollary 4. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n}\left( {n \geq 3}\right) \) be given non-negative numbers, and let \( 0 \leq {x}_{1} \leq {x}_{2} \leq \cdots \leq {x}_{n} \) such that\n\n\[ \n{x}_{1} + {x}_{2} + \cdots + {x}_{n} = {a}_{1} + {a}_{2} + \cdots + {a}_{n},{x}_{1}^{p} + {x}_{2}^{p} + \cdots + {x}_{n}^{p} ... | Proof. Apply EV-Theorem to the function \( f\left( u\right) = p\ln u \) . We see that \( \mathop{\lim }\limits_{{u \rightarrow 0}}f\left( u\right) = - \infty \) for \( p > 0 \), and\n\n\[ \n{f}^{\prime }\left( u\right) = \frac{p}{u}, g\left( x\right) = {f}^{\prime }\left( {x}^{\frac{1}{p - 1}}\right) = p{x}^{\frac{1}{1... | Yes |
Corollary 5. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n}\left( {n \geq 3}\right) \) be given non-negative numbers, and let \( 0 \leq {x}_{1} \leq {x}_{2} \leq \cdot \; \leq {x}_{n} \) such that\n\n\[ \n{x}_{1} + {x}_{2} + \cdots + {x}_{n} = {a}_{1} + {a}_{2} + \cdots + {a}_{n},{x}_{1}^{p} + {x}_{2}^{p} + \cdots + {x}_{n}^{p... | Proof. We will apply EV-Theorem to the function\n\n\[ \nf\left( u\right) = q\left( {q - 1}\right) \left( {q - p}\right) {u}^{q}. \n\]\n\nFor \( p > 0 \), it is easy to check that either \( f\left( u\right) \) is continuous at \( u = 0 \) (in the case \( q > 0 \) ) or \( \mathop{\lim }\limits_{{u \rightarrow 0}}f\left( ... | Yes |
Corollary 6. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n}\left( {n \geq 3}\right) \) be given non-negative numbers, let \( p \in \{ 1,2\} \) and let \( 0 \leq {x}_{1} \leq {x}_{2} \leq \cdot \; \leq {x}_{n} \) such that\n\n\[ \n{x}_{1} + {x}_{2} + \cdots + {x}_{n} = {a}_{1} + {a}_{2} + \cdots + {a}_{n}, \]\n\n\[ \n{x}_{1}^{p... | Proof Taking into account the known relation\n\n\[ \n6\sum {x}_{1}{x}_{2}{x}_{3} = {\left( \sum {x}_{1}\right) }^{3} - 3\left( {\sum {x}_{1}}\right) \left( {\sum {x}_{1}^{2}}\right) + 2\sum {x}_{1}^{3} \]\n\nthe statement follows by Corollary 5 (case \( p = 2 \) and \( q = 3 \), or \( p = 3 \) and \( q = 2) \) | Yes |
Corollary 7. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n}\left( {n \geq 3}\right) \) by given non-negative numbers, and let \( 0 \leq {x}_{1} \leq {x}_{2} \leq . \leq {x}_{n} \) such that\n\n\[ \n{x}_{1}^{2} + {x}_{2}^{2} + \cdots + {x}_{n}^{2} = {a}_{1}^{2} + {a}_{2}^{2} + \cdots + {a}_{n}^{2} \]\n\n\[ \n{x}_{1}^{3} + {x}_{... | Proof. According to the relation\n\n\[ \n6\sum {x}_{1}{x}_{2}{x}_{3} = {\left( \sum {x}_{1}\right) }^{3} - 3\left( {\sum {x}_{1}}\right) \left( {\sum {x}_{1}^{2}}\right) + 2\sum {x}_{1}^{3} \]\n\nthe sum \( \sum {x}_{1}{x}_{2}{x}_{3} \) is maximal (minimal) when \( \sum {x}_{1} \) is maximal (minimal)\n\nConsequently, ... | Yes |
Corollary 4 (case \( p = - 1 \) ) If \( 0 < {a}_{1} \leq {a}_{2} \leq \cdot \leq {a}_{n} \) such that\n\n\[ \n{a}_{1} + {a}_{2} + \cdots + {a}_{n} = n\\text{ and }\\frac{1}{{a}_{1}} + \\frac{1}{{a}_{2}} + \\cdots + \\frac{1}{{a}_{n}} = \\text{ constant,}\n\]\n\nthen the product \( {a}_{1}{a}_{2}\\ldots {a}_{n} \) is ma... | Denoting \( {a}_{1} = x \) and \( {a}_{2} = {a}_{3} = \\cdots = {a}_{n} = y \), we have to prove that for \( 0 < x \leq 1 \leq y < \\frac{n}{n - 1} \) and \( x + \\left( {n - 1}\\right) y = n \), the inequality holds\n\n\[ \n{y}^{n - 1} + \\left( {n - 1}\\right) x{y}^{n - 2} - \\left( {n - {e}_{n - 1}}\\right) x{y}^{n ... | Yes |
Theorem 1 (AM-GM inequality). For all positive real numbers \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \), the following inequality holds\n\n\[ \frac{{a}_{1} + {a}_{2} + \ldots + {a}_{n}}{n} \geq \sqrt[n]{{a}_{1}{a}_{2}\ldots {a}_{n}} \]\n\nEquality occurs if and only if \( {a}_{1} = {a}_{2} = \ldots = {a}_{n} \) . | Proof. The inequality is clearly true for \( n = 2 \) . If it is true for \( n \) numbers, it will be true for \( {2n} \) numbers because\n\n\[ {a}_{1} + {a}_{2} + ... + {a}_{2n} \geq n\sqrt[n]{{a}_{1}{a}_{2}...{a}_{n}} + n\sqrt[n]{{a}_{n + 1}{a}_{n + 2}...{a}_{2n}} \geq {2n}\sqrt[{2n}]{{a}_{1}{a}_{2}...{a}_{n}}, \]\n\... | Yes |
Prove that for all non-negative real numbers \( a, b, c \) ,\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2} \] | Consider the following expressions\n\n\[ S = \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}; \]\n\n\[ M = \frac{b}{b + c} + \frac{c}{c + a} + \frac{a}{a + b}; \]\n\n\[ N = \frac{c}{b + c} + \frac{a}{c + a} + \frac{b}{a + b} \]\n\nWe have of course \( M + N = 3 \) . According to AM-GM, we get\n\n\[ M + S = \frac{a ... | Yes |
Proposition 2 (Weighted AM-GM inequality). Suppose that \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are positive real numbers. If \( n \) non-negative real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) have sum 1 then\n\n\[ \n{a}_{1}{x}_{1} + {a}_{2}{x}_{2} + \ldots + {a}_{n}{x}_{n} \geq {a}_{1}^{{x}_{1}}{a}_{2}^{{x}_{2}}\ld... | Solution. The proof of this inequality is entirely similar to the one for the classical AM-GM inequality. However, in the case \( n = 2 \), we need a more detailed proof (because the inequality is posed for real exponents). We have to prove that if \( x, y \geq \) \( 0, x + y = 1 \) and \( a, b > 0 \) then\n\n\[ \n{ax}... | Yes |
Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \sqrt{a} + \sqrt{b} + \sqrt{c} \geq {ab} + {bc} + {ca}. \]\n | Solution. Notice that\n\n\[ 2\left( {{ab} + {bc} + {ca}}\right) = {\left( a + b + c\right) }^{2} - {a}^{2} + {b}^{2} + {c}^{2}. \]\n\nThe inequality is then equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}{a}^{2} + 2\mathop{\sum }\limits_{{cyc}}\sqrt{a} \geq 9 \]\n\nwhich is true by AM-GM because\n\n\[ \mathop{\sum }\... | Yes |
Let \( x, y, z \) be positive real numbers such that \( {xyz} = 1 \) . Prove that\n\n\[ \frac{{x}^{3}}{\left( {1 + y}\right) \left( {1 + z}\right) } + \frac{{y}^{3}}{\left( {1 + z}\right) \left( {1 + x}\right) } + \frac{{z}^{3}}{\left( {1 + x}\right) \left( {1 + y}\right) } \geq \frac{3}{4}. \] | Solution. We use AM-GM in the following form:\n\n\[ \frac{{x}^{3}}{\left( {1 + y}\right) \left( {1 + z}\right) } + \frac{1 + y}{8} + \frac{1 + z}{8} \geq \frac{3x}{4}. \]\n\nWe conclude that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{x}^{3}}{\left( {1 + y}\right) \left( {1 + z}\right) } + \frac{1}{4}\mathop{\sum }\limit... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \left( {1 + \frac{x}{y}}\right) \left( {1 + \frac{y}{z}}\right) \left( {1 + \frac{z}{x}}\right) \geq 2 + \frac{2\left( {x + y + z}\right) }{\sqrt[3]{xyz}}. \]\n | Solution. Certainly, the problem follows the inequality\n\n\[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x + y + z}{\sqrt[3]{xyz}} \]\n\nwhich is true by AM-GM because\n\n\[ 3\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}\right) = \left( {\frac{2x}{y} + \frac{y}{z}}\right) + \left( {\frac{2y}{z} + \frac{z}{x}... | Yes |
Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \n{16}\left( {{abc} + {bcd} + {cda} + {dab}}\right) \leq {\left( a + b + c + d\right) }^{4}.\n\] | Solution. Applying AM-GM for two numbers, we obtain\n\n\[ \n{16}\left( {{abc} + {bcd} + {cda} + {dab}}\right) = {16ab}\left( {c + d}\right) + {16cd}\left( {a + b}\right)\n\]\n\n\[ \n\leq 4{\left( a + b\right) }^{2}\left( {c + d}\right) + 4{\left( c + d\right) }^{2}\left( {a + b}\right)\n\]\n\n\[ \n= 4\left( {a + b + c ... | Yes |
Suppose that \( a, b, c \) are three side-lengths of a triangle with perimeter 3. Prove that\n\n\[ \frac{1}{\sqrt{a + b - c}} + \frac{1}{\sqrt{b + c - a}} + \frac{1}{\sqrt{c + a - b}} \geq \frac{9}{{ab} + {bc} + {ca}}. \] | Solution. Let \( x = \sqrt{b + c - a}, y = \sqrt{c + a - b}, z = \sqrt{a + b - c} \) . We get \( {x}^{2} + {y}^{2} + {z}^{2} = 3. The inequality becomes\n\n\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{36}{9 + {x}^{2}{y}^{2} + {y}^{2}{z}^{2} + {z}^{2}{x}^{2}}. \]\n\nLet \( m = {xy}, n = {yz}, p = {zx} \) . The ... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers such that \( {a}_{i} \in \left\lbrack {0, i}\right\rbrack \) for all \( i \in \{ 1,2,\ldots, n\} \) . Prove that\n\n\[ \n{2}^{n}{a}_{1}\left( {{a}_{1} + {a}_{2}}\right) \ldots \left( {{a}_{1} + {a}_{2} + \ldots + {a}_{n}}\right) \geq \left( {n + 1}\righ... | Solution. According to AM-GM,\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{k} = 1 \cdot \left( \frac{{a}_{1}}{1}\right) + 2 \cdot \left( \frac{{a}_{2}}{2}\right) + \ldots + k \cdot \left( \frac{{a}_{k}}{k}\right)\n\]\n\n\[ \n\geq \frac{k\left( {k + 1}\right) }{2}{\left( \frac{{a}_{1}}{1}\right) }^{\frac{2}{k\left( {k + 1}... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{1}{{a}^{3} + {b}^{3} + {abc}} + \frac{1}{{b}^{3} + {c}^{3} + {abc}} + \frac{1}{{c}^{3} + {a}^{3} + {abc}} \leq \frac{1}{abc}. \]\n\n(USA MO 1998) | Solution. Notice that \( {a}^{3} + {b}^{3} \geq {ab}\left( {a + b}\right) \), so\n\n\[ \frac{abc}{{a}^{3} + {b}^{3} + {c}^{3}} \leq \frac{abc}{{ab}\left( {a + b}\right) + {abc}} = \frac{c}{a + b + c}. \]\n\nBuilding up two similar inequalities and adding up all of them, we have the conclusion\n\n\[ \frac{abc}{{a}^{3} +... | Yes |
Prove that \( {x}_{1}{x}_{2}\ldots {x}_{n} \geq {\left( n - 1\right) }^{n} \) if \( {x}_{1},{x}_{2},\ldots ,{x}_{n} > 0 \) satisfy\n\n\[ \frac{1}{1 + {x}_{1}} + \frac{1}{1 + {x}_{2}} + \ldots + \frac{1}{1 + {x}_{n}} = 1. \] | Solution. The condition implies that\n\n\[ \frac{1}{1 + {x}_{1}} + \frac{1}{1 + {x}_{2}} + \ldots + \frac{1}{1 + {x}_{n - 1}} = \frac{{x}_{n}}{1 + {x}_{n}}. \]\n\nUsing AM-GM inequality for all terms on the left hand side, we obtain\n\n\[ \frac{{x}_{n}}{1 + {x}_{n}} \geq \frac{n - 1}{\sqrt[{n - 1}]{\left( {1 + {x}_{1}}... | Yes |
Suppose that \( x, y, z \) are positive real numbers and \( {x}^{5} + {y}^{5} + {z}^{5} = 3 \) . Prove that \[ \frac{{x}^{4}}{{y}^{3}} + \frac{{y}^{4}}{{z}^{3}} + \frac{{z}^{4}}{{x}^{3}} \geq 3 \] | Solution. Notice that \[ {\left( {x}^{5} + {y}^{5} + {z}^{5}\right) }^{2} = {x}^{10} + 2{x}^{5}{y}^{5} + {y}^{10} + 2{y}^{5}{z}^{5} + {z}^{10} + 2{z}^{5}{x}^{5} = 9. \] This form suggests the AM-GM inequality in the following form \[ {10} \cdot \frac{{x}^{4}}{{y}^{3}} + 6{x}^{5}{y}^{5} + 3{x}^{10} \geq {19}{x}^{\frac{1... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \sqrt{\frac{a + b}{a + 1}} + \sqrt{\frac{b + c}{b + 1}} + \sqrt{\frac{c + a}{c + 1}} \geq 3. \] | Solution. After applying AM-GM for the three terms on the left hand side expression, we only need to prove that\n\n\[ \left( {a + b}\right) \left( {b + c}\right) \left( {c + a}\right) \geq \left( {a + 1}\right) \left( {b + 1}\right) \left( {c + 1}\right) ,\]\n\nor equivalent by (because \( {abc} = 1 \) )\n\n\[ {ab}\lef... | Yes |
Let \( a, b, c \) be the side-lengths of a triangle. Prove that\n\n\[{\left( a + b - c\right) }^{a}{\left( b + c - a\right) }^{b}{\left( c + a - b\right) }^{c} \leq {a}^{a}{b}^{b}{c}^{c}.\] | Solution. Applying the weighted AM-GM inequality, we conclude that\n\n\[a + b + c\sqrt{{\left( \frac{a + b - c}{a}\right) }^{a}{\left( \frac{b + c - a}{b}\right) }^{b}{\left( \frac{c + a - b}{c}\right) }^{c}}\]\n\n\[\leq \frac{1}{a + b + c}\left( {a \cdot \frac{a + b - c}{a} + b \cdot \frac{b + c - a}{b} + c \cdot \fra... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 2. Prove that\n\n\[ \n{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2} \leq 2.\n\] | Solution. We certainly have\n\n\[ \n\left( {{ab} + {bc} + {ca}}\right) \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \geq \mathop{\sum }\limits_{{cyc}}{a}^{3}\left( {b + c}\right) = \mathop{\sum }\limits_{{cyc}}{ab}\left( {{a}^{2} + {b}^{2}}\right) \geq 2\mathop{\sum }\limits_{{cyc}}{a}^{2}{b}^{2}.\n\]\n\nApplying AM-GM,... | Yes |
Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \frac{1}{{a}^{2} + {ab}} + \frac{1}{{b}^{2} + {bc}} + \frac{1}{{c}^{2} + {cd}} + \frac{1}{{d}^{2} + {da}} \geq \frac{4}{{ac} + {bd}}. \]\n | Solution. Notice that\n\n\[ \frac{{ac} + {bd}}{{a}^{2} + {ab}} = \frac{{a}^{2} + {ab} + {ac} + {bd}}{a\left( {a + b}\right) } - 1 = \frac{a\left( {a + c}\right) + b\left( {d + a}\right) }{a\left( {a + b}\right) } - 1 = \frac{a + c}{a + b} + \frac{b\left( {d + a}\right) }{a\left( {a + d}\right) } - 1. \]\n\nAccording to... | Yes |
Let \( a, b, c, d, e \) be non-negative real numbers such that \( a + b + c + d + e = 5 \) . Prove that\n\n\[ {abc} + {bcd} + {cde} + {dea} + {eab} \leq 5. \] | Solution. Without loss of generality, we may assume that \( e = \min \left( {a, b, c, d, e}\right) \) . According to AM-GM, we have\n\n\[ {abc} + {bcd} + {cde} + {dea} + {eab} = e\left( {a + c}\right) \left( {b + d}\right) + {bc}\left( {a + d - e}\right) \]\n\n\[ \leq e{\left( \frac{a + c + b + d}{2}\right) }^{2} + {\l... | Yes |
Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \n{\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) }^{2} \geq \frac{1}{{a}^{2}} + \frac{4}{{a}^{2} + {b}^{2}} + \frac{9}{{a}^{2} + {b}^{2} + {c}^{2}} + \frac{16}{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}. \n\]\n\n(Pham Kim Hung) | Solution. We have to prove that\n\n\[ \n\frac{1}{{b}^{2}} + \frac{1}{{c}^{2}} + \frac{1}{{d}^{2}} + \mathop{\sum }\limits_{\text{sym }}\frac{2}{ab} \geq \frac{4}{{a}^{2} + {b}^{2}} + \frac{9}{{a}^{2} + {b}^{2} + {c}^{2}} + \frac{16}{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}. \n\]\n\n## 1.0. AM-GM inequality\n\nBy AM-GM, w... | Yes |
Determine the least \( M \) for which the inequality\n\n\[ \left| {{ab}\left( {{a}^{2} - {b}^{2}}\right) + {bc}\left( {{b}^{2} - {c}^{2}}\right) + {ca}\left( {{c}^{2} - {a}^{2}}\right) }\right| \leq M{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2} \]\n\nholds for all real numbers \( a, b \) and \( c \) . | Solution. Denote \( x = a - b, y = b - c, z = c - a \) and \( s = a + b + c \) . Rewrite the inequality in the following form\n\n\[ 9\left| {\operatorname{sx}{yz}}\right| \leq M{\left( {s}^{2} + {x}^{2} + {y}^{2} + {z}^{2}\right) }^{2} \]\n\nin which \( s, x, y, z \) are arbitrary real numbers with \( x + y + z = 0 \) ... | Yes |
Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \frac{a}{1 + {b}^{2}} + \frac{b}{1 + {c}^{2}} + \frac{c}{1 + {a}^{2}} \geq \frac{3}{2}. \] | Solution. In fact, it's impossible to use AM-GM for the denominators because the sign will be reversed\n\n\[ \frac{a}{1 + {b}^{2}} + \frac{b}{1 + {c}^{2}} + \frac{c}{1 + {a}^{2}} \leq \frac{a}{2b} + \frac{b}{2c} + \frac{c}{2a} \geq \frac{3}{2}?! \]\n\nHowever, we can use the same application in another appearance\n\n\[... | Yes |
Suppose that \( a, b, c, d \) are four positive real numbers with sum 4. Prove that\n\n\[ \frac{a}{1 + {b}^{2}c} + \frac{b}{1 + {c}^{2}d} + \frac{c}{1 + {d}^{2}a} + \frac{d}{1 + {a}^{2}b} \geq 2. \] | Solution. According to AM-GM, we deduce that\n\n\[ \frac{a}{1 + {b}^{2}c} = a - \frac{a{b}^{2}c}{1 + {b}^{2}c} \geq a - \frac{a{b}^{2}c}{{2b}\sqrt{c}} = a - \frac{{ab}\sqrt{c}}{2} \]\n\n\[ = a - \frac{b\sqrt{a \cdot {ac}}}{2} \geq a - \frac{b\left( {a + {ac}}\right) }{4}. \]\n\nAccording to this estimation,\n\n\[ \math... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{3}}{{a}^{2} + {b}^{2}} + \frac{{b}^{3}}{{b}^{2} + {c}^{2}} + \frac{{c}^{3}}{{c}^{2} + {d}^{2}} + \frac{{d}^{3}}{{d}^{2} + {a}^{2}} \geq \frac{a + b + c + d}{2}. \] | Solution. We use the following estimation\n\n\[ \frac{{a}^{3}}{{a}^{2} + {b}^{2}} = a - \frac{a{b}^{2}}{{a}^{2} + {b}^{2}} \geq a - \frac{a{b}^{2}}{2ab} = a - \frac{b}{2}. \] | No |
Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \frac{{a}^{2}}{a + 2{b}^{2}} + \frac{{b}^{2}}{b + 2{c}^{2}} + \frac{{c}^{2}}{c + 2{a}^{2}} \geq 1. \] | Solution. We use the following estimation according to AM-GM\n\n\[ \frac{{a}^{2}}{a + 2{b}^{2}} = a - \frac{{2a}{b}^{2}}{a + 2{b}^{2}} \geq a - \frac{{2a}{b}^{2}}{3\sqrt[3]{a{b}^{4}}} = a - \frac{2{\left( ab\right) }^{2/3}}{3}, \]\n\nwhich implies that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{a + 2{b}^{2}} \ge... | Yes |
Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \frac{{a}^{2}}{a + 2{b}^{2}} + \frac{{b}^{2}}{b + 2{c}^{3}} + \frac{{c}^{2}}{c + 2{a}^{3}} \geq 1. \] | Solution. Using the same technique as in example 1.2.4, we only need to prove that\n\n\[ b\sqrt[3]{{a}^{2}} + c\sqrt[3]{{b}^{2}} + a\sqrt[3]{{c}^{2}} \leq 3. \]\n\nAccording to AM-GM, we obtain\n\n\[ 3\mathop{\sum }\limits_{{cyc}}a \geq \mathop{\sum }\limits_{{cyc}}a + 2\mathop{\sum }\limits_{{cyc}}{ab} = \mathop{\sum ... | Yes |
Example 1.2.6. Let \( a, b, c \) be positive real numbers which sum up to 3. Prove that\n\n\[ \frac{a + 1}{{b}^{2} + 1} + \frac{b + 1}{{c}^{2} + 1} + \frac{c + 1}{{a}^{2} + 1} \geq 3. \] | Solution. We use the following estimation\n\n\[ \frac{a + 1}{{b}^{2} + 1} = a + 1 - \frac{{b}^{2}\left( {a + 1}\right) }{{b}^{2} + 1} \geq a + 1 - \frac{{b}^{2}\left( {a + 1}\right) }{2b} = a + 1 - \frac{{ab} + b}{2}. \]\n\nSumming up the similar results for \( a, b, c \), we deduce that\n\n\[ \mathop{\sum }\limits_{{c... | Yes |
Let \( a, b, c \) be positive real numbers with sum 3. Prove that\n\n\[ \frac{1}{1 + 2{b}^{2}c} + \frac{1}{1 + 2{c}^{2}a} + \frac{1}{1 + 2{a}^{2}b} \geq 1. \] | Solution. We use the following estimation\n\n\[ \frac{1}{1 + 2{b}^{2}c} = 1 - \frac{2{b}^{2}c}{1 + 2{b}^{2}c} \geq 1 - \frac{2\sqrt[3]{{b}^{2}c}}{3} \geq 1 - \frac{2\left( {{2b} + c}\right) }{9}. \] | No |
Let \( a, b, c, d \) be non-negative real numbers with sum \( 4. \) Prove that\n\n\[ \frac{1 + {ab}}{1 + {b}^{2}{c}^{2}} + \frac{1 + {bc}}{1 + {c}^{2}{d}^{2}} + \frac{1 + {cd}}{1 + {d}^{2}{a}^{2}} + \frac{1 + {da}}{1 + {a}^{2}{b}^{2}} \geq 4. \] | Solution. Applying AM-GM, we have\n\n\[ \frac{1 + {ab}}{1 + {b}^{2}{c}^{2}} = \left( {1 + {ab}}\right) - \frac{\left( {1 + {ab}}\right) {b}^{2}{c}^{2}}{1 + {b}^{2}{c}^{2}} \geq 1 + {ab} - \frac{1}{2}\left( {1 + {ab}}\right) {bc}. \]\n\nSumming up similar results, we get\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1 + {ab}... | Yes |
Example 1.2.9. Let \( a, b, c \) be positive real numbers satisfying \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \frac{1}{{a}^{3} + 2} + \frac{1}{{b}^{3} + 2} + \frac{1}{{c}^{3} + 2} \geq 1 \] | Solution. According to AM-GM, we obtain\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{{a}^{3} + 2} = \frac{3}{2} - \frac{1}{2}\mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{{a}^{3} + 1 + 1} \]\n\n\[ \geq \frac{3}{2} - \frac{1}{2}\mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{3a} = 1 \] | Yes |
Theorem 2 (Cauchy-Schwarz inequality). Let \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) \) be two sequences of real numbers. We have\n\n\[ \left( {{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2}}\right) \left( {{b}_{1}^{2} + {b}_{2}^{2} + \ldots + {b}_{n}... | First solution.(using quadratic form) Consider the following function\n\n\[ f\left( x\right) = {\left( {a}_{1}x - {b}_{1}\right) }^{2} + {\left( {a}_{2}x - {b}_{2}\right) }^{2} + \ldots + {\left( {a}_{n}x - {b}_{n}\right) }^{2} \]\n\nwhich is rewritten as\n\n\[ f\left( x\right) = \left( {{a}_{1}^{2} + {a}_{2}^{2} + \ld... | Yes |
Corollary 1. (Schwarz inequality). For any two sequences of real numbers \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) ,\left( {{b}_{i} > 0\forall i \in \{ 1,2,\ldots, n\} }\right) \), we have\n\n\[ \frac{{a}_{1}^{2}}{{b}_{1}} + \frac{{a}_{2}^{2}}{{b}_{2}} ... | Solution. This result is directly obtained from Cauchy-Schwarz. | Yes |
For every two sequences of real numbers \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) \), we always have\n\n\[ \sqrt{{a}_{1}^{2} + {b}_{1}^{2}} + \sqrt{{a}_{2}^{2} + {b}_{2}^{2}} + ... + \sqrt{{a}_{n}^{2} + {b}_{n}^{2}} \geq \sqrt{{\left( {a}_{1} + ... + {a... | Proof. By a simple induction, it suffices to prove the problem in the case \( n = 2 \) . In this case, the inequality becomes\n\n\[ \sqrt{{a}_{1}^{2} + {b}_{1}^{2}} + \sqrt{{a}_{2}^{2} + {b}_{2}^{2}} \geq \sqrt{{\left( {a}_{1} + {a}_{2}\right) }^{2} + {\left( {b}_{1} + {b}_{2}\right) }^{2}} \]\n\nSquaring and reducing ... | Yes |
Corollary 3. For any sequence of real numbers \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) we have\n\n\[ \n{\left( {a}_{1} + {a}_{2} + \ldots + {a}_{n}\right) }^{2} \leq n\left( {{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2}}\right) .\n\] | Proof. Use Cauchy-Schwarz for the following sequences of \( n \) terms\n\n\[ \n\left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) ,\;\left( {1,1,\ldots ,1}\right) .\n\] | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{2} - {bc}}{2{a}^{2} + {b}^{2} + {c}^{2}} + \frac{{b}^{2} - {ca}}{2{b}^{2} + {c}^{2} + {a}^{2}} + \frac{{c}^{2} - {ab}}{2{c}^{2} + {a}^{2} + {b}^{2}} \geq 0. \]\n\n(Pham Kim Hung) | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{\left( a + b\right) }^{2}}{{a}^{2} + {b}^{2} + 2{c}^{2}} \leq 3 \]\n\nAccording to Cauchy-Schwarz inequality, we have\n\n\[ \frac{{\left( a + b\right) }^{2}}{{a}^{2} + {b}^{2} + 2{c}^{2}} \leq \frac{{a}^{2}}{{a}^{2} + {c}^{2}} + \frac{{... | Yes |
Suppose that \( x, y, z \geq 1 \) and \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2 \) . Prove that\n\n\[ \sqrt{x + y + z} \geq \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}. \]\n\n(Iran MO 1998) | Solution. By hypothesis, we obtain\n\n\[ \frac{x - 1}{x} + \frac{y - 1}{y} + \frac{z - 1}{z} = 1 \]\n\nAccording to Cauchy-Schwarz, we have\n\n\[ \mathop{\sum }\limits_{{cyc}}x = \left( {\mathop{\sum }\limits_{{cyc}}x}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{x - 1}{x}}\right) \geq {\left( \mathop{\sum }\limit... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{3}}{{a}^{3} + {b}^{3} + {abc}} + \frac{{b}^{3}}{{b}^{3} + {c}^{3} + {abc}} + \frac{{c}^{3}}{{c}^{3} + {a}^{3} + {abc}} \geq 1. \] | Solution. Let \( x = \frac{b}{a}, y = \frac{c}{b} \) and \( z = \frac{a}{c} \) . Then we have\n\n\[ \frac{{a}^{3}}{{a}^{3} + {b}^{3} + {abc}} = \frac{1}{1 + {x}^{3} + \frac{x}{z}} = \frac{1}{1 + {x}^{3} + {x}^{2}z} = \frac{yz}{{yz} + {x}^{2} + {xz}}. \]\n\nBy Cauchy-Schwarz inequality, we deduce that\n\n\[ \mathop{\sum... | Yes |
Let \( a, b, c \) be three arbitrary real numbers. Denote\n\n\[ \nx = \sqrt{{b}^{2} - {bc} + {c}^{2}}, y = \sqrt{{c}^{2} - {ca} + {a}^{2}}, z = \sqrt{{a}^{2} - {ab} + {b}^{2}}.\n\]\n\nProve that\n\n\[ \n{xy} + {yz} + {zx} \geq {a}^{2} + {b}^{2} + {c}^{2}.\n\]\n\n(Nguyen Anh Tuan, VMEO 2006) | Solution. Rewrite \( x, y \) in the following forms\n\n\[ \nx = \sqrt{\frac{3{c}^{2}}{4} + {\left( b - \frac{c}{2}\right) }^{2}}, y = \sqrt{\frac{3{c}^{2}}{4} + {\left( a - \frac{c}{2}\right) }^{2}}.\n\]\n\nAccording to Cauchy-Schwarz inequality, we conclude\n\n\[ \n{xy} \geq \frac{3{c}^{2}}{4} + \frac{1}{4}\left( {{2b... | Yes |
Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{{b}^{2} + {c}^{2} + {d}^{2}} + \frac{b}{{a}^{2} + {c}^{2} + {d}^{2}} + \frac{c}{{a}^{2} + {b}^{2} + {d}^{2}} + \frac{d}{{a}^{2} + {b}^{2} + {c}^{2}} \geq \frac{4}{a + b + c + d}. \]\n\n(Pham Kim Hung) | Solution. According to Cauchy-Schwarz, we have\n\n\[ \left( {\frac{a}{{b}^{2} + {c}^{2} + {d}^{2}} + \frac{b}{{a}^{2} + {c}^{2} + {d}^{2}} + \frac{c}{{a}^{2} + {b}^{2} + {d}^{2}} + \frac{d}{{a}^{2} + {b}^{2} + {c}^{2}}}\right) \left( {a + b + c + d}\right) \]\n\n\[ \geq {\left( \sqrt{\frac{{a}^{2}}{{b}^{2} + {c}^{2} + ... | Yes |
Prove that for all positive real numbers \( a, b, c, d, e, f \) , we always have\n\n\[ \frac{a}{b + c} + \frac{b}{c + d} + \frac{c}{d + e} + \frac{d}{e + f} + \frac{e}{f + a} + \frac{f}{a + b} \geq 3. \] | Solution. According to Cauchy-Schwarz inequality\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a}{b + c} = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{{ab} + {ac}} \geq \frac{{\left( a + b + c + d + e + f\right) }^{2}}{{ab} + {bc} + {cd} + {de} + {ef} + {fa} + {ac} + {ce} + {ea} + {bd} + {df} + {fb}} \]\n\nDenote the denomi... | Yes |
Prove the following inequality\n\n\[ \n{\left( {a}_{1}{b}_{2} - {a}_{2}{b}_{1}\right) }^{2} \leq 2\left| {{a}_{1}{b}_{1} + {a}_{2}{b}_{2} + \ldots + {a}_{n}{b}_{n} - 1}\right| .\n\] | Solution. By Cauchy-Schwarz, the condition \( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i}^{2} = 1 \) yields\n\n\[ \n1 \geq {a}_{1}{b}_{1} + {a}_{2}{b}_{2} + \ldots + {a}_{n}{b}_{n} \geq - 1 \n\]\n\nAccording to the expansion of the Cauchy-Schwarz inequality, we have\n\n\[... | Yes |
Suppose \( a, b, c \) are positive real numbers with sum 3. Prove that\n\n\[ \sqrt{a + \sqrt{{b}^{2} + {c}^{2}}} + \sqrt{b + \sqrt{{c}^{2} + {a}^{2}}} + \sqrt{c + \sqrt{{a}^{2} + {b}^{2}}} \geq 3\sqrt{\sqrt{2} + 1}. \]\n\n(Phan Hong Son) | Solution. We rewrite the inequality in the following form (after squaring both sides)\n\n\[ \mathop{\sum }\limits_{{cyc}}\sqrt{{b}^{2} + {c}^{2}} + 2\mathop{\sum }\limits_{{cyc}}\sqrt{\left( {a + \sqrt{{b}^{2} + {c}^{2}}}\right) \left( {b + \sqrt{{c}^{2} + {a}^{2}}}\right) } \geq 9\sqrt{2} + 6. \]\n\nAccording to Cauch... | Yes |
Suppose \( a, b, c \) are positive real numbers such that \( {abc} = 1 \) . Prove the following inequality\n\n\[ \frac{1}{{a}^{2} + a + 1} + \frac{1}{{b}^{2} + b + 1} + \frac{1}{{c}^{2} + c + 1} \geq 1. \] | Solution. By hypothesis, there exist three positive real numbers \( x, y, z \) for which\n\n\[ a = \frac{yz}{{x}^{2}}, b = \frac{xz}{{y}^{2}}, c = \frac{xy}{{z}^{2}}. \]\n\nThe inequality can be rewritten to\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{x}^{4}}{{x}^{4} + {x}^{2}{yz} + {y}^{2}{z}^{2}} \geq 1 \]\n\nAccording... | Yes |
Let \( a, b, c \) be the side-lengths of a triangle. Prove that\n\n\[ \frac{a}{{3a} - b + c} + \frac{b}{{3b} - c + a} + \frac{c}{{3c} - a + b} \geq 1. \] | Solution. By Cauchy-Schwarz, we have\n\n\[ 4\mathop{\sum }\limits_{{cyc}}\frac{a}{{3a} - b + c} = \mathop{\sum }\limits_{{cyc}}\frac{4a}{{3a} - b + c} \]\n\n\[ = 3 + \mathop{\sum }\limits_{{cyc}}\frac{a + b - c}{{3a} - b + c} \]\n\n\[ \geq 3 + \frac{{\left( a + b + c\right) }^{2}}{\mathop{\sum }\limits_{{cyc}}\left( {a... | Yes |
Let \( a, b, c \) be positive real numbers such that \( a \leq b \leq c \) and \( a + b + c = 3 \) . Prove that\n\n\[ \sqrt{3{a}^{2} + 1} + \sqrt{5{a}^{2} + 3{b}^{2} + 1} + \sqrt{7{a}^{2} + 5{b}^{2} + 3{c}^{2} + 1} \leq 9. \] | Solution. According to Cauchy-Schwarz, we have\n\n\[ {\left( \sqrt{3{a}^{2} + 1} + \sqrt{5{a}^{2} + 3{b}^{2} + 1} + \sqrt{7{a}^{2} + 5{b}^{2} + 3{c}^{2} + 1}\right) }^{2} = \]\n\n\[ = {\left( \frac{1}{\sqrt{6}} \cdot \sqrt{6(3{a}^{2} + 1)} + \frac{1}{\sqrt{4}} \cdot \sqrt{4(5{a}^{2} + 3{b}^{2} + 1)} + \frac{1}{\sqrt{3}... | Yes |
Let \( a, b, c \) be positive real numbers with sum 1. Prove that\n\n\[ \frac{1 + a}{1 - a} + \frac{1 + b}{1 - b} + \frac{1 + c}{1 - c} \leq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a}. \]\n\n(Japan TST 2004) | Solution. Rewrite this inequality in the form\n\n\[ \frac{3}{2} + \mathop{\sum }\limits_{{cyc}}\frac{a}{b + c} \leq \mathop{\sum }\limits_{{cyc}}\frac{a}{b} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\left( {\frac{a}{b} - \frac{a}{b + c}}\right) \geq \frac{3}{2} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{ac}{... | Yes |
Prove that for all non-negative real numbers \( x, y, z \)\n\n\[\n6\left( {x + y - z}\right) \left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) + {27xyz} \leq {10}{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{3/2}.\n\] | For this part, it's necessary to be aware of the fact that the equality holds for \( x = y = {2z} \) up to permutation. This suggests using orientated estimations. Indeed, by Cauchy-Schwarz, we deduce that\n\n\[{10}{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{3/2} - 6\left( {x + y - z}\right) \left( {{x}^{2} + {y}^{2}... | Yes |
Let \( a, b, c, d \) be four positive real numbers such that \( {r}^{4} = {abcd} \geq 1 \). Prove the following inequality \[ \frac{{ab} + 1}{a + 1} + \frac{{bc} + 1}{b + 1} + \frac{{cd} + 1}{c + 1} + \frac{{da} + 1}{d + 1} \geq \frac{4\left( {1 + {r}^{2}}\right) }{1 + r}. \] | The hypothesis implies the existence of four positive real numbers \( x, y, z, t \) such that \[ a = \frac{ry}{x},\;b = \frac{rz}{y},\;c = \frac{rt}{z},\;d = \frac{rx}{t}. \] The inequality is therefore rewritten in the following form \[ \mathop{\sum }\limits_{{cyc}}\frac{\frac{{r}^{2}z}{x} + 1}{\frac{ry}{x} + 1} \geq ... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{2}}{{a}^{2} + 2{\left( a + b\right) }^{2}} + \frac{{b}^{2}}{{b}^{2} + 2{\left( b + c\right) }^{2}} + \frac{{c}^{2}}{{c}^{2} + 2{\left( c + a\right) }^{2}} \geq \frac{1}{3}. \]\n\n(Pham Kim Hung) | Solution. We denote \( x = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c} \). The problem becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{1 + 2{\left( x + 1\right) }^{2}} \geq \frac{1}{3} \]\n\nBecause \( {xyz} = 1 \), there exist three positive real numbers \( m, n, p \) such that\n\n\[ x = \frac{np}{{m}^{2}}, y = ... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{{b}^{2} + {c}^{2}} + \frac{b}{{a}^{2} + {c}^{2}} + \frac{c}{{a}^{2} + {b}^{2}} \geq \frac{4}{5}\left( {\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}}\right) . \]\n\n(Pham Kim Hung) | Solution. Applying Cauchy-Schwarz, we obtain\n\n\[ \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{{b}^{2} + {c}^{2}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}a\left( {{b}^{2} + {c}^{2}}\right) }\right) \geq {\left( a + b + c\right) }^{2}. \]\n\nIt remains to prove that\n\n\[ \frac{{\left( a + b + c\right) }^{2}}{{ab... | Yes |
Corollary 1. Let \( a, b, c, x, y, z, t, u, v \) be positive real numbers. We always have\n\n\[ \left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) \left( {{x}^{3} + {y}^{3} + {z}^{3}}\right) \left( {{t}^{3} + {u}^{3} + {v}^{3}}\right) \geq {\left( axt + byu + czv\right) }^{3}. \] | Proof. This is a direct corollary of Hölder inequality for \( m = n = 3 \) . I choose this particular case of Hölder for a detailed proof because it exemplifies the proof of the general Hölder inequality.\n\nAccording to AM-GM, we deduce that\n\n\[ 3 = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{{a}^{3} + {b}^{3} + {c}... | No |
Corollary 2. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers. Prove that\n\n\[ \left( {1 + {a}_{1}}\right) \left( {1 + {a}_{2}}\right) \ldots \left( {1 + {a}_{n}}\right) \geq {\left( 1 + \sqrt[n]{{a}_{1}{a}_{2}\ldots {a}_{n}}\right) }^{n}. \]\n | Proof. Applying AM-GM, we have\n\n\[ \frac{1}{1 + {a}_{1}} + \frac{1}{1 + {a}_{2}} + \ldots + \frac{1}{1 + {a}_{n}} \geq \frac{n}{\sqrt[n]{\left( {1 + {a}_{1}}\right) \left( {1 + {a}_{2}}\right) \ldots \left( {1 + {a}_{n}}\right) }}, \]\n\n\[ \frac{{a}_{1}}{1 + {a}_{1}} + \frac{{a}_{2}}{1 + {a}_{2}} + \ldots + \frac{{a... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a}{\sqrt{{a}^{2} + {8bc}}} + \frac{b}{\sqrt{{b}^{2} + {8ac}}} + \frac{c}{\sqrt{{c}^{2} + {8ab}}} \geq 1. \]\n\n(IMO 2001, A2) | Solution. Applying Hölder inequality for three sequences, each of which has three terms (actually, that's corollary 1), we deduce that\n\n\[ \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{{a}^{2} + {8bc}}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{{a}^{2} + {8bc}}}}\right) \left( {\mathop{\sum }\... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \frac{a}{\sqrt{7 + b + c}} + \frac{b}{\sqrt{7 + c + a}} + \frac{c}{\sqrt{7 + a + b}} \geq 1 \] | For the first one, apply Hölder inequality in the following form\n\n\[ \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{7 + b + c}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{7 + b + c}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}a(7 + b + c)}\right) \geq (a + b + c{)}^{3} \]\n\nIt's enough to pro... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that for all natural numbers \( k,\left( {k \geq 1}\right) \), the following inequality holds\n\n\[ \frac{{a}^{k + 1}}{{b}^{k}} + \frac{{b}^{k + 1}}{{c}^{k}} + \frac{{c}^{k + 1}}{{a}^{k}} \geq \frac{{a}^{k}}{{b}^{k - 1}} + \frac{{b}^{k}}{{c}^{k - 1}} + \frac{{c}^{k}}{{a... | Solution. According to Hölder inequality, we deduce that\n\n\[ {\left( \frac{{a}^{k + 1}}{{b}^{k}} + \frac{{b}^{k + 1}}{{c}^{k}} + \frac{{c}^{k + 1}}{{a}^{k}}\right) }^{k - 1}\left( {a + b + c}\right) \geq {\left( \frac{{a}^{k}}{{b}^{k - 1}} + \frac{{b}^{k}}{{c}^{k - 1}} + \frac{{c}^{k}}{{a}^{k - 1}}\right) }^{k}. \]\n... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \left( {{a}^{5} - {a}^{2} + 3}\right) \left( {{b}^{5} - {b}^{2} + 3}\right) \left( {{c}^{5} - {c}^{2} + 3}\right) \geq {\left( a + b + c\right) }^{3}. \]\n\n\[ \text{(Titu Andreescu, USA MO 2002)} \] | Solution. According to Hölder inequality, we conclude that\n\n\[ \mathop{\prod }\limits_{{cyc}}\left( {{a}^{5} - {a}^{2} + 3}\right) = \mathop{\prod }\limits_{{cyc}}\left( {{a}^{3} + 2 + \left( {{a}^{3} - 1}\right) \left( {{a}^{2} - 1}\right) }\right) \geq \prod \left( {{a}^{3} + 2}\right) \]\n\n\[ = \left( {{a}^{3} + ... | Yes |
Suppose \( a, b, c \) are three positive real numbers verifying \( {ab} + {bc} + {ca} = 3\). Prove that \[ \left( {1 + {a}^{2}}\right) \left( {1 + {b}^{2}}\right) \left( {1 + {c}^{2}}\right) \geq 8 \] | The inequality is directly obtained from Hölder inequality \[ \left( {{a}^{2}{b}^{2} + {a}^{2} + {b}^{2} + 1}\right) \left( {{b}^{2} + {c}^{2} + {b}^{2}{c}^{2} + 1}\right) \left( {{a}^{2} + {a}^{2}{c}^{2} + {c}^{2} + 1}\right) \geq {\left( 1 + ab + bc + ca\right) }^{4}. \] | Yes |
Let \( a, b, c \) be positive real numbers which sum up to 1 . Prove that\n\n\[\n\frac{a}{\sqrt[3]{a + {2b}}} + \frac{b}{\sqrt[3]{b + {2c}}} + \frac{c}{\sqrt[3]{c + {2a}}} \geq 1.\n\]\n\n(Pham Kim Hung) | Solution. This inequality is directly obtained from by Hölder inequality\n\n\[\n\left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt[3]{a + {2b}}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt[3]{a + {2b}}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt[3]{a + {2b}}}}\right) \left( {\mathop{\s... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n{a}^{2}\left( {b + c}\right) + {b}^{2}\left( {c + a}\right) + {c}^{2}\left( {a + b}\right) \geq \left( {{ab} + {bc} + {ca}}\right) \sqrt[3]{\left( {a + b}\right) \left( {b + c}\right) \left( {c + a}\right) }.\n\]\n\n(Pham Kim Hung) | Solution. Notice that the following expressions are equal to each other\n\n\[ \n{a}^{2}\left( {b + c}\right) + {b}^{2}\left( {c + a}\right) + {c}^{2}\left( {a + b}\right)\n\]\n\n\[ \n{b}^{2}\left( {c + a}\right) + {c}^{2}\left( {a + b}\right) + {a}^{2}\left( {b + c}\right)\n\]\n\n\[ \n{ab}\left( {a + b}\right) + {bc}\l... | Yes |
Let \( a, b, c, d \) be positive real numbers such that \( {abcd} = 1 \) . Prove that\n\n\[ \n{4}^{4}\left( {{a}^{4} + 1}\right) \left( {{b}^{4} + 1}\right) \left( {{c}^{4} + 1}\right) \left( {{d}^{4} + 1}\right) \geq {\left( a + b + c + d + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) }^{4}.\n\]\n\n(Ga... | Solution. By Hölder inequality, we get that\n\n\[ \n\left( {{a}^{4} + 1}\right) \left( {1 + {b}^{4}}\right) \left( {1 + {c}^{4}}\right) \left( {1 + {d}^{4}}\right) \geq {\left( a + bcd\right) }^{4} = {\left( a + \frac{1}{a}\right) }^{4}\n\]\n\n\[ \n\Rightarrow \sqrt[4]{\left( {{a}^{4} + 1}\right) \left( {{b}^{4} + 1}\r... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \left( {{a}^{2} + {ab} + {b}^{2}}\right) \left( {{b}^{2} + {bc} + {c}^{2}}\right) \left( {{c}^{2} + {ca} + {a}^{2}}\right) \geq {\left( ab + bc + ca\right) }^{3}. \]\n | Solution. Applying Hölder inequality, we obtain\n\n\[ \left( {{a}^{2} + {ab} + {b}^{2}}\right) \left( {{b}^{2} + {bc} + {c}^{2}}\right) \left( {{c}^{2} + {ca} + {a}^{2}}\right) \]\n\n\[ = \left( {{ab} + {a}^{2} + {b}^{2}}\right) \left( {{a}^{2} + {ac} + {c}^{2}}\right) \left( {{b}^{2} + {c}^{2} + {bc}}\right) \geq {\le... | Yes |
Example 2.2.10. Suppose that \( a, b, c \) are positive real numbers satisfying the condition \( 3\max \left( {{a}^{2},{b}^{2},{c}^{2}}\right) \leq 2\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \) . Prove that\n\n\[ \frac{a}{\sqrt{2{b}^{2} + 2{c}^{2} - {a}^{2}}} + \frac{b}{\sqrt{2{c}^{2} + 2{a}^{2} - {b}^{2}}} + \frac{c... | Solution. By Hölder, we deduce that\n\n\[ \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{2{b}^{2} + 2{c}^{2} - {a}^{2}}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{\sqrt{2{b}^{2} + 2{c}^{2} - {a}^{2}}}}\right) \left( {\mathop{\sum }\limits_{{cyc}}a\left( {2{b}^{2} + 2{c}^{2} - {a}^{2}}\right) }\right) \... | Yes |
Theorem 4 (Chebyshev inequality). Suppose \( \\left( {{a}_{1},{a}_{2},\\ldots ,{a}_{n}}\\right) \) and \( \\left( {{b}_{1},{b}_{2},\\ldots ,{b}_{n}}\\right) \) are two increasing sequences of real numbers, then\n\n\[ \n{a}_{1}{b}_{1} + {a}_{2}{b}_{2} + \\ldots + {a}_{n}{b}_{n} \\geq \\frac{1}{n}\\left( {{a}_{1} + {a}_{... | Proof. By directly expanding, we have\n\n\[ \nn\\left( {{a}_{1}{b}_{1} + {a}_{2}{b}_{2} + \\ldots + {a}_{n}{b}_{n}}\\right) - \\left( {{a}_{1} + {a}_{2} + \\ldots + {a}_{n}}\\right) \\left( {{b}_{1} + {b}_{2} + \\ldots + {b}_{n}}\\right) =\n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{{i, j = 1}}^{n}\\left( {{a}_{i} - {a}_{j... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers with sum \( n \) . Prove that\n\n\[ {a}_{1}^{n + 1} + {a}_{2}^{n + 1} + \ldots + {a}_{n}^{n + 1} \geq {a}_{1}^{n} + {a}_{2}^{n} + \ldots + {a}_{n}^{n}. \]\n | Solution. To solve this problem by AM-GM, we must go through two steps: first, prove \( n\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{n + 1} + n \geq \left( {n + 1}\right) \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{n} \), and then prove \( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{n} \geq n \) . To solve it by Cauchy... | No |
Let \( a, b, c, d \) be positive real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = 4 \) . Prove the following inequality\n\n\[ \frac{{a}^{2}}{b + c + d} + \frac{{b}^{2}}{c + d + a} + \frac{{c}^{2}}{d + a + b} + \frac{{d}^{2}}{a + b + c} \geq \frac{4}{3}. \] | Solution. Notice that if \( \left( {a, b, c, d}\right) \) is arranged in an increasing order then\n\n\[ \frac{1}{b + c + d} \geq \frac{1}{c + d + a} \geq \frac{1}{d + a + b} \geq \frac{1}{a + b + c}. \]\n\nTherefore, by Chebyshev inequality, we have\n\n\[ 4\mathrm{{LHS}} \geq \left( {\mathop{\sum }\limits_{{cyc}}{a}^{2... | Yes |
Suppose that the real numbers \( a, b, c > 1 \) satisfy the condition\n\n\[ \frac{1}{{a}^{2} - 1} + \frac{1}{{b}^{2} - 1} + \frac{1}{{c}^{2} - 1} = 1 \]\n\nProve that\n\n\[ \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \leq 1 \] | Solution. Notice that if \( a \geq b \geq c \) then we have\n\n\[ \frac{a - 2}{a + 1} \geq \frac{b - 2}{b + 1} \geq \frac{c - 2}{c + 1}\;;\;\frac{a + 2}{a - 1} \leq \frac{b + 2}{b - 1} \leq \frac{c + 2}{c - 1}. \]\n\nChebyshev inequality affirms that\n\n\[ 3\left( {\mathop{\sum }\limits_{{cyc}}\frac{{a}^{2} - 4}{{a}^{2... | Yes |
Let \( a, b, c, d, e \) be non-negative real numbers such that\n\n\[ \frac{1}{4 + a} + \frac{1}{4 + b} + \frac{1}{4 + c} + \frac{1}{4 + d} + \frac{1}{4 + e} = 1. \]\n\nProve that\n\n\[ \frac{a}{4 + {a}^{2}} + \frac{b}{4 + {b}^{2}} + \frac{c}{4 + {c}^{2}} + \frac{d}{4 + {d}^{2}} + \frac{e}{4 + {e}^{2}} \leq 1. \] | Solution. The hypothesis implies that \( \mathop{\sum }\limits_{\text{cyc }}\frac{1 - a}{4 + a} = 0 \) . We need to prove that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{4 + a} \geq \mathop{\sum }\limits_{{cyc}}\frac{a}{4 + {a}^{2}} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{1 - a}{4 + a} \cdot \frac{1}{4 + {a... | Yes |
Example 3.1.5. Suppose that \( a, b, c, d \) are four positive real numbers satisfying \( a + b + c + d = 4 \) . Prove that\n\n\[ \frac{1}{{11} + {a}^{2}} + \frac{1}{{11} + {b}^{2}} + \frac{1}{{11} + {c}^{2}} + \frac{1}{{11} + {d}^{2}} \leq \frac{1}{3}. \]\n | Solution. Rewrite the inequality in the following form\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{1}{{11} + {a}^{2}} - \frac{1}{12}}\right) \geq 0 \]\n\nor equivalently\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {1 - a}\right) \cdot \frac{a + 1}{{a}^{2} + {11}} \geq 0 \]\n\nNotice that if \( \left( {a, b, c, d}\ri... | Yes |
Let \( a, b, c \) be three positive real numbers with sum 3. Prove that\n\n\[ \frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{c}^{2}} \geq {a}^{2} + {b}^{2} + {c}^{2} \] | Solution. Rewrite the inequality in the form\n\n\[ \mathop{\sum }\limits_{{cyc}}{a}^{2}{b}^{2} \geq {a}^{2}{b}^{2}{c}^{2}\mathop{\sum }\limits_{{cyc}}{a}^{2} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}{a}^{2}{b}^{2}\left( {1 + c + {c}^{2} + {c}^{3}}\right) \left( {1 - c}\right) \geq 0. \]\n\nNotice that if \( {ab} \l... | Yes |
Example 3.2.1. Suppose \( a, b, c, d \) are positive real numbers such that\n\n\[ a + b + c + d = {a}^{-1} + {b}^{-1} + {c}^{-1} + {d}^{-1}. \]\n\nProve the inequality\n\n\[ 2\left( {a + b + c + d}\right) \geq \sqrt{{a}^{2} + 3} + \sqrt{{b}^{2} + 3} + \sqrt{{c}^{2} + 3} + \sqrt{{d}^{2} + 3}. \] | Solution. A cursory look at this inequality will leave you hesitating. The relationship between the variables \( a, b, c, d \) appears to be obscure and very hard to transform; moreover, the problem involves square roots. How can use handle this situation? Surprisingly enough, a simple way of applying Chebyshev can dra... | Yes |
Example 3.2.2. Suppose \( a, b, c \) are positive real numbers with sum 3. Prove that\n\n\[ \frac{1}{{c}^{2} + a + b} + \frac{1}{{a}^{2} + b + c} + \frac{1}{{b}^{2} + a + c} \leq 1. \] | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{1}{{c}^{2} - c + 3} - \frac{1}{3}}\right) \geq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\left( \frac{a\left( {a - 1}\right) }{{a}^{2} - a + 3}\right) \geq 0 \]\n\nor\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( \frac{a - 1}{a ... | Yes |
Example 3.2.3. Let \( a, b, c \) be positive real numbers and \( 0 \leq k \leq 2 \) . Prove that\n\n\[ \n\frac{{a}^{2} - {bc}}{{b}^{2} + {c}^{2} + k{a}^{2}} + \frac{{b}^{2} - {ca}}{{c}^{2} + {a}^{2} + k{b}^{2}} + \frac{{c}^{2} - {ab}}{{a}^{2} + {b}^{2} + k{c}^{2}} \geq 0.\n\]\n\n(Pham Kim Hung) | Solution. Although this problem can be solved in the same way as example 2.1.1 is solved, we can use Chebyshev inequality to give a simpler solution. Notice that if \( a \geq b \) then for all positive real \( c \), we have \( \left( {{a}^{2} - {bc}}\right) \left( {b + c}\right) \geq \left( {{b}^{2} - {ca}}\right) \lef... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \sqrt{{a}^{2} + {8bc}} + \sqrt{{b}^{2} + {8ca}} + \sqrt{{c}^{2} + {8ab}} \leq 3\left( {a + b + c}\right) . \] | Solution. Rewrite the inequality in the following form\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {{3a} - \sqrt{{a}^{2} + {8bc}}}\right) \geq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2} - {bc}}{{3a} + \sqrt{{a}^{2} + {8bc}}} \geq 0 \]\n\nor\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{\left( {{a}^{2} - {bc}... | Yes |
Let \( a, b, c, d \) be positive real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = 4 \) . Prove that\n\n\[ \frac{1}{5 - a} + \frac{1}{5 - b} + \frac{1}{5 - c} + \frac{1}{5 - d} \leq 1 \] | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{1}{5 - a} - \frac{1}{4}}\right) \leq 0 \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{a - 1}{5 - a} \leq 0 \]\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{\left( {a - 1}\right) \left( {a + 1}\right) }{\left( {5 - ... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers satisfying\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{n} = \frac{1}{{a}_{1}} + \frac{1}{{a}_{2}} + \ldots + \frac{1}{{a}_{n}}.\n\]\n\nProve that the following inequality holds\n\n\[ \n\frac{1}{{n}^{2} + {a}_{1}^{2} - 1} + \frac{1}{{n}^{2} + {a}_{2}^{2} -... | Solution. WLOG, we may assume that \( {a}_{1} \geq {a}_{2} \geq \ldots \geq {a}_{n} \) . The hypothesis is equivalent to:\n\n\[ \n\frac{1 - {a}_{1}^{2}}{{a}_{1}} + \frac{1 - {a}_{2}^{2}}{{a}_{2}} + \ldots + \frac{1 - {a}_{n}^{2}}{{a}_{n}} = 0\left( *\right)\n\]\n\nDenote \( S = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i... | Yes |
Suppose that \( a, b, c \) are positive real numbers with sum 3. Prove that\n\n\[ \frac{1}{9 - {ab}} + \frac{1}{9 - {bc}} + \frac{1}{9 - {ca}} \leq \frac{3}{8}. \] | Solution. Let \( x = {bc}, y = {ca}, z = {ab} \) . The inequality becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{9 - x} \leq \frac{3}{8} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{1 - x}{9 - x} \geq 0. \]\n\nSuppose that \( {a}_{x},{a}_{y},{a}_{z} \) are the coefficients we are looking for. We will rewrite... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {a}^{4} + {b}^{4} + {c}^{4} = 3 \) . Prove that \[ \frac{1}{4 - {ab}} + \frac{1}{4 - {bc}} + \frac{1}{4 - {ca}} \leq 1 \] | Solution. Let \( x = {ab}, y = {ac} \) and \( z = {bc} \) . The inequality is equivalent to \[ \frac{1 - x}{4 - x} + \frac{1 - y}{4 - y} + \frac{1 - z}{4 - z} \geq 0 \] \[ \Leftrightarrow \frac{1 - {x}^{2}}{4 + {3x} - {x}^{2}} + \frac{1 - {y}^{2}}{4 + {3y} - {y}^{2}} + \frac{1 - {z}^{2}}{4 + {3z} - {z}^{2}} \geq 0 \] N... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers such that\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{n} = \frac{1}{{a}_{1}} + \frac{1}{{a}_{2}} + \ldots + \frac{1}{{a}_{n}}.\n\]\n\nProve the following inequality\n\n\[ \n\frac{1}{n - 1 + {a}_{1}^{2}} + \frac{1}{n - 1 + {a}_{2}^{2}} + \ldots + \frac{1}{... | Solution. Rewrite the inequality to the following from\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}\left( {\frac{1}{n - 1 + {a}_{i}^{2}} - \frac{1}{n}}\right) \leq 0\n\]\n\nor equivalently\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}\frac{{a}_{i}^{2} - 1}{n - 1 + {a}_{i}^{2}} \geq 0\n\]\n\nAssume that \( {a}_{1} \geq {a}... | Yes |
Theorem 5. If \( f\left( x\right) \) is a real function defined on \( \left\lbrack {a, b}\right\rbrack \subset \mathbb{R} \) and \( {f}^{\prime \prime }\left( x\right) \geq 0\forall x \in \left\lbrack {a, b}\right\rbrack \) then \( f\left( x\right) \) is a convex function on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. We will prove that for all \( x, y \in \left\lbrack {a, b}\right\rbrack \) and for all \( 0 \leq t \leq 1 \)\n\n\[ \n{tf}\left( x\right) + \left( {1 - t}\right) f\left( y\right) \geq f\left( {{tx} + \left( {1 - t}\right) y}\right) .\n\]\n\nIndeed, suppose that \( t \) and \( y \) are constant. Denote\n\n\[ \ng\l... | Yes |
Lemma 1. Suppose that a real function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) satisfies the condition\n\n\[ f\left( x\right) + f\left( y\right) \geq {2f}\left( \frac{x + y}{2}\right) \forall x, y \in \left\lbrack {a, b}\right\rbrack ,\]\n\nthen for all \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \in \l... | Proof. We use Cauchy induction to solve this lemma. By hypothesis, the inequality holds for \( n = 2 \), therefore it holds for every number \( n \) that is a power of 2 . It’s enough to prove that if the inequality holds for \( n = k + 1 - \left( {k \in \mathbb{N}, k \geq 2}\right) \) then it will hold for \( n = k \)... | Yes |
Lemma 2. Suppose that the real function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{ + } \) satisfies the condition\n\n\[ f\left( x\right) + f\left( y\right) \geq {2f}\left( \sqrt{xy}\right) \forall x, y \in \left\lbrack {a, b}\right\rbrack \]\n\nthen for all \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \in... | The proof of this lemma is completely similar to that of lemma 1 and therefore it won't be shown here. | No |
Lemma 3. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be non-negative real numbers with sum 1 and \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be real numbers in \( \left\lbrack {a, b}\right\rbrack \) . Let \( f\left( x\right) \) be a real function defined on \( \left\lbrack {a, b}\right\rbrack \) . The inequality\n\n\[ \n{a}_{1... | To prove lemma 3 as well as the weighted Jensen inequality, we use the same method as in the proof of lemma 1. | No |
Suppose that \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) are positive real numbers and \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \geq 1 \) . Prove that\n\n\[ \frac{1}{1 + {x}_{1}} + \frac{1}{1 + {x}_{2}} + \ldots + \frac{1}{1 + {x}_{n}} \leq \frac{n}{1 + \sqrt[n]{{x}_{1}{x}_{2}\ldots {x}_{n}}}. \]\n\n(IMO Shortlist) | Solution. According to lema 2, it's enough to prove that\n\n\[ \frac{1}{1 + {a}^{2}} + \frac{1}{1 + {b}^{2}} \leq \frac{2}{1 + {ab}}\forall a, b \geq 1. \]\n\nWe can reduce this inequality to \( {\left( a - b\right) }^{2}\left( {1 - {ab}}\right) \leq 0 \), which is obvious. | No |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be real numbers lying in \( (1/2,1\rbrack \) . Prove that\n\n\[ \frac{{a}_{1}{a}_{2}\ldots {a}_{n}}{{\left( {a}_{1} + {a}_{2} + \ldots + {a}_{n}\right) }^{n}} \geq \frac{\left( {1 - {a}_{1}}\right) \left( {1 - {a}_{2}}\right) \ldots \left( {1 - {a}_{n}}\right) }{{\left( n - {a}... | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\left( {\ln {a}_{i} - \ln \left( {1 - {a}_{i}}\right) }\right) \geq n\ln \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}}\right) - n\ln \left( {n - \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}}\right) .\n\nNotice that the function \( f\... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a}{\sqrt{{a}^{2} + {8bc}}} + \frac{b}{\sqrt{{b}^{2} + {8ac}}} + \frac{c}{\sqrt{{c}^{2} + {8ab}}} \geq 1. \]\n\n(IMO 2001, A2) | Solution. Although this problem has been solved using Hölder, a proof by Jensen's inequality is very nice, too. WLOG, we may assume that \( a + b + c = 1 \) normalize. Because \( f\left( x\right) = \frac{1}{\sqrt{x}} \) is a convex function, we obtain from Jensen’s inequality that:\n\n\[ a \cdot f\left( {{a}^{2} + {8bc... | Yes |
Let \( a, b, c, d \) be positive numbers with sum 4. Prove that\n\n\[ \frac{a}{{b}^{2} + b} + \frac{b}{{c}^{2} + c} + \frac{c}{{d}^{2} + d} + \frac{d}{{a}^{2} + a} \geq \frac{8}{\left( {a + c}\right) \left( {b + d}\right) }. \]\n | Solution. Denote \( f\left( x\right) = \frac{1}{x\left( {x + 1}\right) } \), then \( f \) is a convex function if \( x > 0 \) . According to Jensen inequality, we have\n\n\[ \frac{a}{4} \cdot f\left( b\right) + \frac{b}{4} \cdot f\left( c\right) + \frac{c}{4} \cdot f\left( d\right) + \frac{d}{4} \cdot f\left( a\right) ... | Yes |
Suppose that \( a, b, c \) are positive real numbers. Prove that\n\n\[ \sqrt{\frac{a}{a + b}} + \sqrt{\frac{b}{b + c}} + \sqrt{\frac{c}{c + a}} \leq \frac{3}{\sqrt{2}}. \] | Solution. Notice that \( f\left( x\right) = \sqrt{x} \) is a concave function. According to Jensen inequality, we have\n\n\[ \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{a}{a + b}} = \mathop{\sum }\limits_{{cyc}}\frac{a + c}{2\left( {a + b + c}\right) } \cdot \sqrt{\frac{{4a}{\left( a + b + c\right) }^{2}}{\left( {a + b}\r... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{\sqrt{4{b}^{2} + {bc} + 4{c}^{2}}} + \frac{b}{\sqrt{4{c}^{2} + {ca} + 4{a}^{2}}} + \frac{c}{\sqrt{4{a}^{2} + {ab} + 4{b}^{2}}} \geq 1. \] | Solution. We may assume that \( a + b + c = 1 \) . Because \( f\left( x\right) = \frac{1}{\sqrt{x}} \) is a convex function, according to Jensen inequality, we have\n\n\[ a \cdot f\left( {4{b}^{2} + {bc} + 4{c}^{2}}\right) + b \cdot f\left( {4{c}^{2} + {ca} + 4{a}^{2}}\right) + c \cdot f\left( {4{a}^{2} + {ab} + 4{b}^{... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \sqrt{\frac{a}{{4a} + {4b} + c}} + \sqrt{\frac{b}{{4b} + {4c} + a}} + \sqrt{\frac{c}{{4c} + {4a} + b}} \leq 1. \] | Solution. Notice that \( f\left( x\right) = \sqrt{x} \) is a concave function, therefore by Jensen inequality we have\n\n\[ \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{a}{{4a} + {4b} + c}} = \mathop{\sum }\limits_{{cyc}}\frac{\left( 4a + 4c + b\right) }{9\left( {a + b + c}\right) } \cdot \sqrt{\frac{{81a}{\left( a + b + c... | Yes |
Example 4.1.8. Let \( a, b, c \) be positive real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that \[ \sqrt{\frac{a}{{a}^{2} + {b}^{2} + 1}} + \sqrt{\frac{b}{{b}^{2} + {c}^{2} + 1}} + \sqrt{\frac{c}{{c}^{2} + {a}^{2} + 1}} \leq \sqrt{3}. \] (Pham Kim Hung) | Solution. Applying Jensen inequality for the concave function \( f\left( x\right) = \sqrt{x} \), we have \[ \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{a}{{a}^{2} + {b}^{2} + 1}} = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2} + {c}^{2} + 1}{3({a}^{2} + {b}^{2} + {c}^{2})} \cdot \sqrt{\frac{{9a}({a}^{2} + {b}^{2} + {c}^{2}{)... | Yes |
Example 4.2.1. Suppose that \( a, b, c \) are positive real numbers belonging to \( \left\lbrack {1,2}\right\rbrack \) . Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} \leq {5abc} \n\]\n\n(MYM 2001) | Solution. Let's first give an elementary solution to this simple problem. Since \( a, b, c \in \left\lbrack {1,2}\right\rbrack \), if \( a \geq b \geq c \) then\n\n\[ \n{a}^{3} + 2 \leq {5a} \Leftrightarrow \left( {a - 2}\right) \left( {{a}^{2} + {2a} - 1}\right) \leq 0 \n\]\n\n(1)\n\n\[ \n{5a} + {b}^{3} \leq {5ab} + 1... | Yes |
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