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Suppose that \( a, b, c \) are positive real numbers belonging to \( \\left\\lbrack {1,2}\\right\\rbrack \) . Prove that\n\n\[ \n\\left( {a + b + c}\\right) \\left( {\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}}\\right) \\leq {10}.\n\]\n\n(Olympiad 30-4, Vietnam) | Solution. The inequality can be rewritten as\n\n\[ \n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + \\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c} \\leq 7.\n\]\n\nWLOG, we may assume that \( a \\geq b \\geq c \), then\n\n\[ \n\\left( {a - b}\\right) \\left( {a - c}\\right) \\geq 0 \\Rightarrow \\left\\{ \\begin{array}{l} \\f... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{2005} \) be real numbers belonging to \( \left\lbrack {-1,1}\right\rbrack \) . Find the minimum value for the following expression\n\n\[ P = {x}_{1}{x}_{2} + {x}_{2}{x}_{3} + \ldots + {x}_{2004}{x}_{2005} + {x}_{2005}{x}_{1}. \] | Solution. Because this inequality is cyclic, not symmetric, we can not order variables as. If we rely on the relation \( \left( {{x}_{i} - 1}\right) \left( {{x}_{i} + 1}\right) \leq 0 \), we won’t succeed either.\n\nBy intuition, we feel that the expression will attain its maximum if in the sequence \( \left( {{x}_{1},... | Yes |
Lemma 4. Suppose that \( F\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) is a real function defined on \( \left\lbrack {a, b}\right\rbrack \times \left\lbrack {a, b}\right\rbrack \times \) \( \cdots \times \left\lbrack {a, b}\right\rbrack \subset {\mathbb{R}}^{n}\left( {a < b}\right) \) such that for all \( k \in \... | Solution. In fact, we only need to prove that if \( f\left( x\right) \) is a real convex function defined on \( \left\lbrack {a, b}\right\rbrack \) then for all \( x \in \left\lbrack {a, b}\right\rbrack \), we have\n\n\[ f\left( x\right) \leq \max \{ f\left( a\right), f\left( b\right) \} . \]\n\nIndeed, since \( \{ {ta... | Yes |
Given positive real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \in \left\lbrack {a, b}\right\rbrack \), find the maximum value of\n\n\[ \n{\left( {x}_{1} - {x}_{2}\right) }^{2} + {\left( {x}_{1} - {x}_{3}\right) }^{2} + \cdots + {\left( {x}_{1} - {x}_{n}\right) }^{2} + {\left( {x}_{2} - {x}_{3}\right) }^{2} + \cdots + ... | Solution. Denote the above expression by \( F \) . Notice that \( F \), represented as a function of \( {x}_{1} \) (we have already fixed other variables), is equal to\n\n\[ \nf\left( {x}_{1}\right) = \left( {n - 1}\right) {x}_{1}^{2} - 2\left( {\mathop{\sum }\limits_{{i = 2}}^{n}{x}_{i}}\right) {x}_{1} + c \n\]\n\nin ... | Yes |
Let \( n \in \mathbb{N} \). Find the minimum value of the following expression \[ f\left( x\right) = \left| {1 + x}\right| + \left| {2 + x}\right| + \ldots + \left| {n + x}\right| ,\;\left( {x \in \mathbb{R}}\right) . | Solution. Denote \( {I}_{1} = \lbrack - 1, + \infty ),{I}_{n + 1} = ( - \infty , - n\rbrack \) and \( {I}_{k} = \left\lbrack {-k, - k + 1}\right\rbrack \) for each \( k \in \{ 2,3,\ldots, n\} \) . If \( x \in {I}_{1} \) then \( f\left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {1 + x}\right) \geq \mathop{\su... | Yes |
Lemma 5. Suppose that \( f\\left( x\\right) \) is a real convex function defined on \( \\left\\lbrack {a, b}\\right\\rbrack \\in \\mathbb{R} \) and \( {x}_{1},{x}_{2},\\ldots ,{x}_{n} \\in \\left\\lbrack {a, b}\\right\\rbrack \) such that \( {x}_{1} + {x}_{2} + \\ldots + {x}_{n} = s = \) constant \( \\left( {{na} \\leq... | Solution. Notice that this lemma can be obtained directly from its case \( n = 2 \). In fact, it’s sufficient to prove that: if \( x, y \\in \\left\\lbrack {a, b}\\right\\rbrack \) and \( {2a} \\leq x + y = s \\leq {2b} \) then\n\n\[ f\\left( x\\right) + f\\left( y\\right) \\leq \\left\\{ \\begin{array}{l} f\\left( a\\... | No |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers belonging to \( \left\lbrack {0,2}\right\rbrack \) such that \( {a}_{1} + {a}_{2} + \ldots + {a}_{n} = n \) . Find the maximum value of\n\n\[ S = {a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2} \] | Solution. Applying the above lemma to the convex function \( f\left( x\right) = {x}^{2} \), we get that \( S \) attains the maximum if and only if \( k \) numbers are equal to 2 and \( n - k - 1 \) numbers are equal to 0 . In this case, we have \( S = {4k} + {\left( n - 2k\right) }^{2} \) . Because \( {a}_{1},{a}_{2},\... | Yes |
Example 4.2.7. Suppose that \( a, b, c \in \left\lbrack {0,2}\right\rbrack \) and \( a + b + c = 5 \) . Prove that\n\n\[{a}^{2} + {b}^{2} + {c}^{2} \leq 9\] | Solution. Suppose that \( a \leq b \leq c \) . According to lemma 5, we deduce that \( {a}^{2} + {b}^{2} + {c}^{2} \) attains the maximum if and only if \( a = 0 \) or \( b = c = 2 \) . The first case \( a = 0 \) is rejected because so, \( 4 \geq b + c = 5 \) is a contradiction. In the second case, we have \( a = 1 \) ... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{2007} \) be real numbers in \( \left\lbrack {-1,1}\right\rbrack \) such that \( {a}_{1} + {a}_{2} + \ldots + {a}_{2007} = 0 \) . Prove that\n\n\[ \n{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{2007}^{2} \leq {2006}.\n\] | Solution. Applying lemma 5, we deduce that the expression \( {a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{2007}^{2} \) attains the maximum if and only if \( k \) numbers equal 1 and \( n - k - 1 \) numbers equal -1. \( k \) must be 1003 and the last number must be 0, so we deduce that\n\n\[ \n{a}_{1}^{2} + {a}_{2}^{2} + \... | No |
Theorem 8 (Abel formula). Suppose that \( \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) and \( \left( {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) \) are two sequences of real numbers. Denote \( {c}_{k} = {y}_{1} + {y}_{2} + \ldots + {y}_{k}\left( {k = 1,2,\ldots, n}\right) \), then\n\n\( {x}_{1}{y}_{1} + {x}_{2}{y}_{... | Proof. We certainly have\n\n\[ \left( {{x}_{1} - {x}_{2}}\right) {c}_{1} + \left( {{x}_{2} - {x}_{3}}\right) {c}_{2} + \ldots + \left( {{x}_{n - 1} - {x}_{n}}\right) {c}_{n - 1} + {x}_{n}{c}_{n} \]\n\n\[ = {c}_{1}{x}_{1} + \left( {{c}_{2} - {c}_{1}}\right) {x}_{2} + \ldots + \left( {{c}_{n} - {c}_{n - 1}}\right) {x}_{n... | Yes |
Example 5.1.1 (Abel inequality). Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) and \( {y}_{1} \geq {y}_{2} \geq \ldots \geq {y}_{n} \geq 0 \) be real numbers. For eack \( k \in \{ 1,2,\ldots, n\} \), we denote \( {S}_{k} = \mathop{\sum }\limits_{{i = 1}}^{k}{x}_{i} \) . Suppose that \( M = \) \( \max \left\{ {{S}_{1},{S}_{... | Solution. Since both two parts of the inequality can be proved similarly, we only need to show the solution to the left inequality. Let \( {y}_{n + 1} = 0 \) . By Abel formula\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}{y}_{i} = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{y}_{i} - {y}_{i + 1}}\right) {S}_{i} \geq ... | Yes |
Let \( {a}_{1},{a}_{2},..,{a}_{n} \) and \( {b}_{1} \geq {b}_{2} \geq \ldots \geq {b}_{n} \geq 0 \) be positive real numbers such that \( {a}_{1}{a}_{2}\ldots {a}_{k} \geq {b}_{1}{b}_{2}\ldots {b}_{k}\forall k \in \{ 1,2,\ldots, n\} \) . Prove the following inequality\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{n} \geq {... | Solution. By Abel formula, we deduce that\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} - \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i} = \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i}\left( {\frac{{a}_{i}}{{b}_{i}} - 1}\right) \n\]\n\n\[ \n= \left( {{b}_{1} - {b}_{2}}\right) \left( {\frac{{a}_{1}}{{b}_{1}} - 1}\right) +... | Yes |
Let \( {x}_{1},{x}_{2},..,{x}_{n} \) be positive real numbers such that\n\n\[ \n{x}_{1} + {x}_{2} + \ldots + {x}_{k} \geq \sqrt{k}\forall k \in \{ 1,2,\ldots, n\} .\n\]\n\nProve the following inequality\n\n\[ \n{x}_{1}^{2} + {x}_{2}^{2} + \ldots + {x}_{n}^{2} \geq \frac{1}{4}\left( {1 + \frac{1}{2} + \frac{1}{3} + \ldo... | Solution. WLOG, assume that \( {x}_{1} \geq {x}_{2} \geq \ldots \geq {x}_{n} \) . For each \( k \in \{ 1,2,\ldots, n\} \), let \( {b}_{k} = \frac{1}{\sqrt{k}} \) . We will first prove that\n\n\[ \n2\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}^{2} \geq \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}{b}_{i}\n\]\n\nand then, \( ... | Yes |
Example 5.1.4. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) and \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) be real numbers such that\n\n\[ \n{a}_{1} \geq \frac{{a}_{1} + {a}_{2}}{2} \geq \ldots \geq \frac{{a}_{1} + {a}_{2} + \ldots + {a}_{n}}{n} \n\]\n\n\[ \n{b}_{1} \geq \frac{{b}_{1} + {b}_{2}}{2} \geq \ldots \geq \frac{{b}_{... | Solution. For each \( k \in \{ 1,2,\ldots, n\} \), we denote \( {S}_{k} = {a}_{1} + {a}_{2} + \ldots + {a}_{k} \) and \( {b}_{n + 1} = 0 \) . By Abel formula, we have\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{b}_{i} - {b}_{i + 1}}\right) {S}_{i} = \mathop{\s... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be real numbers such that \( {x}_{1} \geq {x}_{2} \geq \ldots \geq {x}_{n} \geq \) \( {x}_{n + 1} = 0 \) . Prove the following inequality\n\n\[ \sqrt{{x}_{1} + {x}_{2} + \ldots + {x}_{n}} \leq \mathop{\sum }\limits_{{i = 1}}^{n}\sqrt{i}\left( {\sqrt{{x}_{i}} - \sqrt{{x}_{i + 1}... | Solution. Denote \( {c}_{i} = \sqrt{i} - \sqrt{i - 1} \) and \( {a}_{i} = \sqrt{{x}_{i}} \) . The inequality becomes\n\n\[ {\left( {a}_{1}{c}_{1} + {a}_{2}{c}_{2} + \ldots + {a}_{n}{c}_{n}\right) }^{2} \geq {a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n}^{2}. \]\n\nSuppose that \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) are po... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) and \( {b}_{1} \leq {b}_{2} \leq \ldots \leq {b}_{n} \) be real numbers such that \( {a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{k}^{2} \leq {b}_{1}^{2} + {b}_{2}^{2} + \ldots + {b}_{k}^{2}\forall k \in \{ 1,2,\ldots, n\} \) . Prove that\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{... | Solution. We prove this problem by induction. Case \( n = 1 \) is obvious. Suppose that the problem has been proved for \( n \) numbers already. We will prove it for \( n + 1 \) numbers. Indeed, by Cauchy-Schwarz, we deduce that\n\n\[ \n\left( {{a}_{1}^{2} + {a}_{2}^{2} + \ldots + {a}_{n + 1}^{2}}\right) \left( {{b}_{1... | Yes |
Let \( - 1 < {x}_{1} < {x}_{2} < \ldots < {x}_{n} < 1 \) and \( {y}_{1} < {y}_{2} < \ldots < {y}_{n} \) be real numbers such that \( {x}_{1} + {x}_{2} + \ldots + {x}_{n} = {x}_{1}^{13} + {x}_{2}^{13} + \ldots + {x}_{n}^{13} \) . Prove that\n\n\[ \n{x}_{1}^{13}{y}_{1} + {x}_{2}^{13}{y}_{2} + \ldots + {x}_{n}^{13}{y}_{n}... | Solution. According to Abel formula, we note that\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}{y}_{i}\left( {{x}_{i}^{13} - {x}_{i}}\right) = \left( {{y}_{1} - {y}_{2}}\right) \left( {{x}_{1}^{13} - {x}_{1}}\right) + \left( {{y}_{2} - {y}_{3}}\right) \left( {{x}_{1}^{13} + {x}_{2}^{13} - {x}_{1} - {x}_{2}}\right) + \ldo... | Yes |
Theorem 9 (Rearrangement Inequality). Let \( \\left( {{a}_{1},{a}_{2},\\ldots ,{a}_{n}}\\right) \) and \( \\left( {{b}_{1},{b}_{2},\\ldots ,{b}_{n}}\\right) \) be two increasing sequences of real numbers. Suppose that \( \\left( {{i}_{1},{i}_{2},\\ldots ,{i}_{n}}\\right) \) is an arbitrary permutation of \( \\left( {1,... | Proof. Notice that \( {a}_{1} \\leq {a}_{2} \\leq \\ldots \\leq {a}_{n} \) and \( {b}_{1} \\leq {b}_{2} \\leq \\ldots \\leq {b}_{n} \), so according to Abel formula,\n\n\[ \n\\mathop{\\sum }\\limits_{{k = 1}}^{n}{a}_{k}{b}_{k} - \\mathop{\\sum }\\limits_{{k = 1}}^{n}{a}_{k}{b}_{{i}_{k}} = \\mathop{\\sum }\\limits_{{k =... | Yes |
Let \( {a}_{1},{a}_{2},..,{a}_{n} \) be positive real numbers. Prove that\n\n\[ \n{a}_{1} + {a}_{2} + \ldots + {a}_{n} \geq n\sqrt[n]{{a}_{1}{a}_{2}\ldots {a}_{n}}. \n\] | Solution. WLOG, assume that \( {a}_{1}{a}_{2}\ldots {a}_{n} = 1 \) (normalization). Let \( {a}_{1} = \frac{{x}_{1}}{{x}_{2}},{a}_{2} = \) \( \frac{{x}_{2}}{{x}_{3}},\ldots ,{a}_{n - 1} = \frac{{x}_{n - 1}}{{x}_{n}},{x}_{1},{x}_{2},\ldots ,{x}_{n} > 0 \), then \( {a}_{n} = \frac{{x}_{n}}{{x}_{1}} \) . The problem become... | Yes |
Suppose that \( a, b, c \) are the side-lengths of a triangle. Prove that\n\n\[ \n{a}^{2}b\left( {a - b}\right) + {b}^{2}c\left( {b - c}\right) + {c}^{2}a\left( {c - a}\right) \geq 0.\n\]\n\n(IMO 1984, A3) | Solution. Because \( a, b, c \) are the side-lengths of a triangle, \( a \geq b \) implies \( {a}^{2} + {bc} \geq \) \( {b}^{2} + {ca} \) . By this property, we deduce that if \( a \geq b \geq c \) then \( {a}^{2} + {bc} \geq {b}^{2} + {ca} \geq {c}^{2} + {ab} \) ; also, \( \frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{2} + {bc}}{b + c} + \frac{{b}^{2} + {ca}}{c + a} + \frac{{c}^{2} + {ab}}{a + b} \geq a + b + c. \] | Solution. Applying Rearrangement inequality for the sequences \( \left( {{a}^{2},{b}^{2},{c}^{2}}\right) \) and \( \left( {\frac{1}{b + c},\frac{1}{c + a},\frac{1}{a + b}}\right) \) (if \( a \geq b \geq c \), then these are both increasing), we get that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{b + c} \geq \mat... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a + b}{a + c} + \frac{a + c}{b + c} + \frac{b + c}{a + b} \leq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}. \]\n\n(Mathlinks Contest) | Solution. The inequality can be rewritten to\n\n\[ \mathop{\sum }\limits_{{cyc}}\left( {\frac{a}{b} - \frac{a}{b + c}}\right) \geq \mathop{\sum }\limits_{{cyc}}\frac{a}{a + c} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{ac}{b\left( {b + c}\right) } \geq \mathop{\sum }\limits_{{cyc}}\frac{a}{a + c}. \]\n\nConside... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a + b}{b + c} + \frac{b + c}{c + a} + \frac{c + a}{a + b} \leq \frac{{\left( a + b + c\right) }^{2}}{{ab} + {bc} + {ca}}. \] | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{\left( {a + b}\right) \left( {a\left( {b + c}\right) + {bc}}\right) }{b + c} \leq {\left( a + b + c\right) }^{2} \]\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{{cyc}}a\left( {a + b}\right) + \mathop{\sum }\limits_{{cyc}}\frac{{bc}\left(... | Yes |
Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \n{a}^{2}{bc} + {b}^{2}{cd} + {c}^{2}{da} + {d}^{2}{ab} \leq 4.\n\] | Solution. Suppose that \( \left( {x, y, z, t}\right) \) is a permutation of \( \left( {a, b, c, d}\right) \) such that \( x \geq y \geq \) \( z \geq t \), then \( {xyz} \geq {xyt} \geq {xzt} \geq {yzt} \) . By Rearrangement inequality, we deduce that\n\n\[ \nx \cdot {xyz} + y \cdot {xyt} + z \cdot {xzt} + t \cdot {yzt}... | Yes |
Example 5.2.7. Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \n{\left( \frac{a}{a + b + c}\right) }^{2} + {\left( \frac{b}{b + c + d}\right) }^{2} + {\left( \frac{c}{c + d + a}\right) }^{2} + {\left( \frac{d}{d + a + b}\right) }^{2} \geq \frac{4}{9}.\n\] | Solution. WLOG, we may assume that \( a + b + c + d = 1 \) . Suppose that \( \left( {x, y, z, t}\right) \) is a permutation of \( \left( {a, b, c, d}\right) \) such that \( x \geq y \geq z \geq t \), then\n\n\[ \n\frac{1}{x + y + z} \leq \frac{1}{x + y + t} \leq \frac{1}{x + z + t} \leq \frac{1}{y + z + t}.\n\]\n\nBy R... | Yes |
Let \( x, y, z, t \) be real numbers satisfying \( {xy} + {yz} + {zt} + {tx} = 1 \) . Find the minimum of the expression\n\n\[ 5{x}^{2} + 4{y}^{2} + 5{z}^{2} + {t}^{2} \] | Solution. We choose a positive number \( l < 5 \) and apply AM-GM inequality\n\n\[ l{x}^{2} + 2{y}^{2} \geq 2\sqrt{2l}{xy} \]\n\n\[ 2{y}^{2} + l{z}^{2} \geq 2\sqrt{2l}{yz} \]\n\n\[ \left( {5 - l}\right) {z}^{2} + 1/2{t}^{2} \geq \sqrt{2\left( {5 - l}\right) }{zt} \]\n\n\[ 1/2{t}^{2} + \left( {5 - l}\right) {x}^{2} \geq... | Yes |
Let \( x, y, z \) be positive real numbers with sum 3. Find the minimum of the expression \( {x}^{2} + {y}^{2} + {z}^{3} \) . | Solution. Let \( a \) and \( b \) be positive real numbers. By AM-GM inequality, we have\n\n\[ \n{x}^{2} + {a}^{2} \geq {2ax} \n\]\n\n\[ \n{y}^{2} + {a}^{2} \geq {2ay} \n\]\n\n\[ \n{z}^{3} + {b}^{3} + {b}^{3} \geq 3{b}^{2}z \n\]\n\nCombining these results yields that \( {x}^{2} + {y}^{2} + {z}^{3} + 2\left( {{a}^{2} + ... | Yes |
Let \( a, b, c \) be three positive constants and \( x, y, z \) three positive variables such that \( {ax} + {by} + {cz} = {xyz} \). Prove that if there exists a unique positive number \( d \) such that\n\n\[ \frac{2}{d} = \frac{1}{a + d} + \frac{1}{b + d} + \frac{1}{c + d} \]\n\nthen the minimum of \( x + y + z \) is\... | Solution. To avoid the complicated condition \( {ax} + {by} + {cz} = {xyz} \), we will find the minimum of the following homogeneous expression\n\n\[ \frac{\left( {{ax} + {by} + {cz}}\right) {\left( x + y + z\right) }^{2}}{xyz}. \]\n\nCertainly, if the minimum of the above expression is equal to \( C \) then the minimu... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive real numbers. Prove that\n\n\[ \n{x}_{1} + \sqrt{{x}_{1}{x}_{2}} + \ldots + \sqrt[n]{{x}_{1}{x}_{2}\ldots {x}_{n}} \leq e\left( {{x}_{1} + {x}_{2} + \ldots + {x}_{n}}\right) .\n\] | Solution. Suppose that \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are positive real numbers. According to AM-GM inequality, we have\n\n\[ \n\sqrt[k]{\left( {{a}_{1}{x}_{1}}\right) \cdot \left( {{a}_{2}{x}_{2}}\right) \cdots \left( {{a}_{k}{x}_{k}}\right) } \leq \frac{{a}_{1}{x}_{1} + {a}_{2}{x}_{2} + \ldots + {a}_{k}{x}_{k}... | Yes |
Let \( x, y, z \) be positive real numbers and \( {xy} + {yz} + {zx} = 1 \) . Prove that\n\n\[ \n{10}{x}^{2} + {10}{y}^{2} + {z}^{2} \geq 4 \n\] | Solution. Let's first see a nice, short and a bit magic solution, before we give a general and natural solution. By AM-GM inequality, we deduce that \( 2{x}^{2} + 2{y}^{2} \geq \) \( {4xy},8{x}^{2} + 1/2{z}^{2} \geq {4xz} \) and \( 8{y}^{2} + 1/2{z}^{2} \geq {4yz} \) . Adding up these inequalities, we have\n\n\[ \n{10}... | Yes |
Let \( k \) be a positive real number. Find the minimum of\n\n\[ k\left( {{x}^{2} + {y}^{2}}\right) + {z}^{2} \]\n\nwhere \( x, y, z \) are three positive real numbers such that \( {xy} + {yz} + {zx} = 1 \) . | Solution. We separate \( k = l + \left( {k - l}\right) \) (with the condition \( 0 \leq l \leq k \) ) and apply AM-GM inequality in the following form\n\n\[ l{x}^{2} + l{y}^{2} \geq {2lxy} \]\n\n\[ \left( {k - l}\right) {x}^{2} + 1/2{z}^{2} \geq \sqrt{2\left( {k - l}\right) }{xz} \]\n\n\[ \left( {k - l}\right) {y}^{2} ... | No |
Suppose that \( x, y, z \) are three positive real numbers verifying \( x + y + z = 3 \). Find the minimum of the expression \[ {x}^{4} + 2{y}^{4} + 3{z}^{4} \] | Let \( a, b, c \) be three positive real numbers such that \( a + b + c = 3 \). According to Hölder inequality, we obtain \[ \left( {{x}^{4} + 2{y}^{4} + 3{z}^{4}}\right) {\left( {a}^{4} + 2{b}^{4} + 3{c}^{4}\right) }^{3} \geq {\left( {a}^{3}x + 2{b}^{3}y + 3{c}^{3}z\right) }^{4}\left( \star \right) \] We will choose \... | Yes |
Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers. Prove that\n\n\[ \frac{1}{{a}_{1}} + \frac{2}{{a}_{1} + {a}_{2}} + \ldots + \frac{n}{{a}_{1} + {a}_{2} + \ldots + {a}_{n}} < 2\left( {\frac{1}{{a}_{1}} + \frac{1}{{a}_{2}} + \ldots + \frac{1}{{a}_{n}}}\right) . \] | Solution. Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive real numbers (we will determine them at the end). According to Cauchy-Schwarz inequality, we have\n\n\[ \left( {{a}_{1} + {a}_{2} + \ldots + {a}_{k}}\right) \left( {\frac{{x}_{1}^{2}}{{a}_{1}} + \frac{{x}_{2}^{2}}{{a}_{2}} + \ldots + \frac{{x}_{k}^{2}}{{a}... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive real numbers. Prove that\n\n\[ \n{x}_{1}^{2} + {\left( \frac{{x}_{1} + {x}_{2}}{2}\right) }^{2} + \ldots + {\left( \frac{{x}_{1} + {x}_{2} + \ldots + {x}_{n}}{n}\right) }^{2} \leq 4\left( {{x}_{1}^{2} + {x}_{2}^{2} + \ldots + {x}_{n}^{2}}\right) .\n\] | Solution. Let \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) be positive real numbers. According to Cauchy-Schwarz inequality, we have\n\n\[ \n\left( {\frac{{x}_{1}^{2}}{{\alpha }_{1}} + \frac{{x}_{2}^{2}}{{\alpha }_{2}} + \ldots + \frac{{x}_{k}^{2}}{{\alpha }_{k}}}\right) \left( {{\alpha }_{1} + {\alpha }_{2}... | Yes |
Find the best value of \( t = t\left( n\right) \) (smallest) for which the following inequality is true for all real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \)\n\n\[ \n{x}_{1}^{2} + {\left( {x}_{1} + {x}_{2}\right) }^{2} + \ldots + {\left( {x}_{1} + {x}_{2} + \ldots + {x}_{n}\right) }^{2} \leq t\left( {{x}_{1}^{2} +... | Solution. Let \( {c}_{1},{c}_{2},\ldots ,{c}_{n} \) be positive real numbers (which will be chosen later). According to Cauchy-Schwarz inequality, we have\n\n\[ \n{\left( \mathop{\sum }\limits_{{i = 1}}^{k}{x}_{i}\right) }^{2} \leq {S}_{k}\left( {\mathop{\sum }\limits_{{i = 1}}^{k}\frac{{x}_{i}^{2}}{{c}_{i}}}\right) \n... | Yes |
Find the minimum value of \( {x}^{x} \), if \( x \) is a positive real number. | Consider the function \( f\left( x\right) = {x}^{x} = {e}^{x\ln x} \) . Its derivative is \( {f}^{\prime }\left( x\right) = \) \( {e}^{x\ln x}\left( {\ln x + 1}\right) \) . Clearly, \( {f}^{\prime }\left( x\right) = 0 \Leftrightarrow \ln x = - 1 \Leftrightarrow x = 1/e \) . In \( \left( {0,\frac{1}{e}}\right\rbrack, f\... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{3}}{{b}^{3} + {c}^{3}} + \frac{{b}^{3}}{{c}^{3} + {a}^{3}} + \frac{{c}^{3}}{{a}^{3} + {b}^{3}} \geq \frac{{a}^{2}}{{b}^{2} + {c}^{2}} + \frac{{b}^{2}}{{c}^{2} + {a}^{2}} + \frac{{c}^{2}}{{a}^{2} + {b}^{2}}. \] | Solution. We will solve the general problem for all real numbers \( s \geq t \geq 0 \)\n\n\[ \frac{{a}^{s}}{{b}^{s} + {c}^{s}} + \frac{{b}^{s}}{{c}^{s} + {a}^{s}} + \frac{{c}^{s}}{{a}^{s} + {b}^{s}} \geq \frac{{a}^{t}}{{b}^{t} + {c}^{t}} + \frac{{b}^{t}}{{c}^{t} + {a}^{t}} + \frac{{c}^{t}}{{a}^{t} + {b}^{t}}. \]\n\nIt'... | Yes |
Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \sqrt{\frac{{ab} + {ac} + {ad} + {bc} + {bd} + {cd}}{6}} \geq \sqrt[3]{\frac{{abc} + {bcd} + {cda} + {dab}}{4}}. \]\n | Solution. Consider the function\n\n\[ f\left( x\right) = \left( {x - a}\right) \left( {x - b}\right) \left( {x - c}\right) \left( {x - d}\right) = {x}^{4} - A{x}^{3} + B{x}^{2} - {Cx} + D \]\n\nwhere\n\n\[ A = \mathop{\sum }\limits_{{sym}}a,\;B = \mathop{\sum }\limits_{{sym}}{ab},\;C = \mathop{\sum }\limits_{{sym}}{abc... | Yes |
Prove that \( 0 \leq a \leq 1 \leq b \leq 3 \leq c \leq 4 \) if \( a, b, c \) are real numbers satisfying the conditions\n\n\[ \na \leq b \leq c, a + b + c = 6,{ab} + {bc} + {ca} = 9.\n\]\n\n(British MO) | Solution. Denote \( p = {abc} \) and consider the function\n\n\[ \nf\left( x\right) = \left( {x - a}\right) \left( {x - b}\right) \left( {x - c}\right) = {x}^{3} - 6{x}^{2} + {9x} - p.\n\]\n\nWe have \( {f}^{\prime }\left( x\right) = 3{x}^{2} - {12x} + 9 = 3\left( {x - 1}\right) \left( {x - 3}\right) \) . Therefore \( ... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {3abc} \geq {ab}\left( {a + b}\right) + {bc}\left( {b + c}\right) + {ca}\left( {c + a}\right) .\n\] | Solution. WLOG, assume that \( a \geq b \geq c \) . Consider the function of \( a : f\left( a\right) = \) \( {a}^{3} + {b}^{3} + {c}^{3} + {3abc} - {ab}\left( {a + b}\right) - {bc}\left( {b + c}\right) - {ca}\left( {c + a}\right) \) . We have\n\n\[ \n{f}^{\prime }\left( a\right) = 3{a}^{2} + {3bc} - {2ab} - {b}^{2} - {... | Yes |
Let \( a, b, c, d \) be positive real numbers such that\n\n\[ 2\left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) + {abc} + {bcd} + {cda} + {dab} = {16}. \]\n\nProve the following inequality\n\n\[ a + b + c + d \geq \frac{2}{3}\left( {{ab} + {bc} + {cd} + {da} + {ac} + {bd}}\right) . \] | Solution. By a similar reasoning as in example 7.1.3, we deduce that there exist positive real numbers \( x, y, z \) for which\n\n\[ \mathop{\sum }\limits_{{sym}}a = \frac{4}{3}\mathop{\sum }\limits_{{sym}}x,\;\mathop{\sum }\limits_{{sym}}{ab} = 2\mathop{\sum }\limits_{{sym}}{xy},\;\mathop{\sum }\limits_{{sym}}{abc} = ... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {9abc} + 4\left( {a + b + c}\right) \geq 8\left( {{ab} + {bc} + {ca}}\right) .\n\]\n\n(Le Trung Kien) | Solution. We denote\n\n\[ \nf\left( b\right) = {b}^{3} + b\left( {4 + {9ac} - {8a} - {8c}}\right) + {a}^{3} + {c}^{3} + 4\left( {a + c}\right) - {8ac}.\n\]\n\nBy AM-GM inequality, \( \left( {{a}^{3} + {4a}}\right) + \left( {{c}^{3} + {4c}}\right) \geq 4{a}^{2} + 4{c}^{2} \geq {8ac} \), so the problem is\n\nproved in ca... | Yes |
Suppose \( n \) is an integer greater than 2 and that the \( n \) positive real numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) satisfy the condition\n\n\[ \left( {{x}_{1} + {x}_{2} + \ldots + {x}_{n}}\right) \left( {\frac{1}{{x}_{1}} + \frac{1}{{x}_{2}} + \ldots + \frac{1}{{x}_{n}}}\right) \leq {\left( n + \sqrt{10} - 3... | Solution. It suffices to prove the following result (that directly solves this problem)\n\nSuppose that \( {x}_{1} \geq {x}_{2} \geq \ldots \geq {x}_{n} > 0 \) are real numbers verifying \( {x}_{1} > {x}_{2} + {x}_{3} \), then\n\n\[ \left( {{x}_{1} + {x}_{2} + \ldots + {x}_{n}}\right) \left( {\frac{1}{{x}_{1}} + \frac{... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {12} \geq {21ab} + {2bc} + {8ca} \) . Prove that\n\n\[\n\frac{1}{a} + \frac{2}{b} + \frac{3}{c} \leq \frac{15}{2}\n\] | Solution. Although this problem has been solved in the previous part by balancing coefficients, it’s seem to be more intuitive to give a solution by derivatives. Let \( x = \) \( \frac{1}{a}, y = \frac{2}{b}, z = \frac{3}{c} \) . We will prove an equivalent problem as follows.\n\n\[\n\text{If}x, y, z > 0\text{and}{12xy... | Yes |
Theorem 10 (Schur inequality). Suppose that \( a, b, c \) are non-negative real numbers, then\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {3abc} \geq {ab}\left( {a + b}\right) + {bc}\left( {b + c}\right) + {ca}\left( {c + a}\right) .\n\] | Proof. Because of the symmetry, we can assume that \( a \geq b \geq c \) . Let \( x = a - b, y = \) \( b - c \) and rewrite the inequality to the following form\n\n\[ \n\mathop{\sum }\limits_{{cyc}}a\left( {a - b}\right) \left( {a - c}\right) \geq 0 \Leftrightarrow c\left( {x + y}\right) y - \left( {c + y}\right) {xy} ... | Yes |
Theorem 11 (General Schur inequality). Suppose that \( a, b, c \) are non-negative real numbers and \( k \) is a positive constant, then\n\n\[ \n{a}^{k}\left( {a - b}\right) \left( {a - c}\right) + {b}^{k}\left( {b - a}\right) \left( {b - c}\right) + {c}^{k}\left( {c - a}\right) \left( {c - b}\right) \geq 0.\n\] | Proof. Certainly, we may assume that \( a \geq b \geq c \) . In this case, we have\n\n\[ \n{c}^{k}\left( {c - a}\right) \left( {c - b}\right) \geq 0\n\]\n\n\[ \n{a}^{k}\left( {a - b}\right) \left( {a - c}\right) + {b}^{k}\left( {b - a}\right) \left( {b - c}\right) = \left( {a - b}\right) \left( {\left( {{a}^{k + 1} - {... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 2. Prove that\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} + {abc} \geq {a}^{3} + {b}^{3} + {c}^{3}.\n\] | Solution. According to the fourth degree-Schur inequality, we have\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} + {abc}\left( {a + b + c}\right) \geq {a}^{3}\left( {b + c}\right) + {b}^{3}\left( {c + a}\right) + {c}^{3}\left( {a + b}\right)\n\]\n\nor equivalently\n\n\[ \n2\left( {{a}^{4} + {b}^{4} + {c}^{4}}\right) + {abc}\left... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{{a}^{2}}{\sqrt{(b + c)({b}^{8} + {c}^{3})}} + \frac{{b}^{2}}{\sqrt{(c + a)({c}^{8} + {a}^{8})}} + \frac{{c}^{2}}{\sqrt{(a + b)({a}^{8} + {b}^{3})}} \geq \frac{3}{2}. \]\n | Solution. By Hölder inequality, we deduce that\n\n\[ {\left( \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{\sqrt{\left( {b + c}\right) \left( {{b}^{3} + {c}^{3}}\right) }}\right) }^{2}\left( {\mathop{\sum }\limits_{{cyc}}{a}^{2}\left( {b + c}\right) \left( {{b}^{3} + {c}^{3}}\right) }\right) \geq {\left( \mathop{\sum }\l... | Yes |
Suppose that \( a, b, c \) are non-negative real numbers. Prove that\n\n\[ \frac{{a}^{2}}{2{b}^{2} - {bc} + 2{c}^{2}} + \frac{{b}^{2}}{2{c}^{2} - {ca} + 2{a}^{2}} + \frac{{c}^{2}}{2{a}^{2} - {ab} + 2{b}^{2}} \geq 1. \] | Solution. According to Cauchy-Schwarz inequality, we deduce that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{2}}{2{b}^{2} - {bc} + 2{c}^{2}} \geq \frac{{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2}}{{a}^{2}\left( {2{b}^{2} - {bc} + 2{c}^{2}}\right) + {b}^{2}\left( {2{c}^{2} - {ca} + 2{a}^{2}}\right) + {c}^{2}\left... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{3}}{{b}^{2} - {bc} + {c}^{2}} + \frac{{b}^{3}}{{c}^{2} - {ca} + {a}^{2}} + \frac{{c}^{3}}{{a}^{2} - {ab} + {b}^{2}} \geq a + b + c. \] | Solution. Applying Cauchy-Schwarz inequality, we have\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{{b}^{2} - {bc} + {c}^{2}} = \mathop{\sum }\limits_{{cyc}}\frac{{a}^{4}}{a\left( {{b}^{2} - {bc} + {c}^{2}}\right) } \geq \frac{{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2}}{\mathop{\sum }\limits_{{cyc}}a\left( {{... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n{a}^{2}\sqrt{{b}^{2} - {bc} + {c}^{2}} + {b}^{2}\sqrt{{c}^{2} - {ca} + {a}^{2}} + {c}^{2}\sqrt{{a}^{2} - {ab} + {b}^{2}} \leq {a}^{3} + {b}^{3} + {c}^{3}.\n\] | Solution. Applying AM-GM inequality, we have\n\n\[ \n\mathop{\sum }\limits_{{cyc}}{a}^{2}\sqrt{{b}^{2} - {bc} + {c}^{2}} = a\sqrt{{a}^{2}\left( {{b}^{2} - {bc} + {c}^{2}}\right) } \leq \frac{1}{2}\mathop{\sum }\limits_{{cyc}}a\left( {{a}^{2} + {b}^{2} + {c}^{2} - {bc}}\right) .\n\]\n\nThen, by the third degree-Schur in... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{a}^{3}}{\sqrt{{b}^{2} - {bc} + {c}^{2}}} + \frac{{b}^{3}}{\sqrt{{c}^{2} - {ca} + {a}^{2}}} + \frac{{c}^{3}}{\sqrt{{a}^{2} - {ab} + {b}^{2}}} \geq {a}^{2} + {b}^{2} + {c}^{2}. \]\n(Vo Quoc Ba Can) | Solution. Applying Cauchy-Schwarz inequality, we deduce that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{a}^{3}}{\sqrt{{b}^{2} - {bc} + {c}^{2}}} \geq \frac{{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2}}{a\sqrt{{b}^{2} - {bc} + {c}^{2}} + b\sqrt{{c}^{2} - {ca} + {a}^{2}} + c\sqrt{{a}^{2} - {ab} + {b}^{2}}}. \]\n\nSo i... | Yes |
Let \( a, b, c \) be positive real numbers with sum 2. Prove that\n\n\[ \n{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2} + {abc} \leq 1.\n\] | Solution. Because \( a + b + c = 2 \), the inequality is equivalent to\n\n\[ \n{\\left( ab + bc + ca\\right) }^{2} \leq 1 + {3abc}.\n\]\n\nDenote \( x = {ab} + {bc} + {ca} \) and \( y = {abc} \) . We are done if \( x \leq 1 \) . Otherwise, \( x \geq 1 \), and by AM-GM inequality, we deduce that\n\n\[ \n\\mathop{\\prod ... | Yes |
Suppose that \( a, b, c \) are non-negative real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} = 1 \) . Prove the inequality\n\n\[ a + b + c \leq \sqrt{2} + \frac{9abc}{4}. \] | Solution. We denote \( x = a + b + c, y = {ab} + {bc} + {ca} \) and \( z = {abc} \) . By the fourth degree-Schur inequality, we have\n\n\[ \mathop{\sum }\limits_{{cyc}}{a}^{4} + {abc}\mathop{\sum }\limits_{{cyc}}a \geq \mathop{\sum }\limits_{{cyc}}{a}^{3}\left( {b + c}\right) \]\n\n\[ \Leftrightarrow {\left( \mathop{\s... | Yes |
Let \( a, b, c \) be positive real numbers satisfying \( {abc} = 1 \) . Prove that\n\n\[ \frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq \frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c}. \]\n\n(Bulgarian MO 1998) | Solution. Denote \( S = \mathop{\sum }\limits_{{cyc}}a, P = \mathop{\sum }\limits_{{cyc}}{ab}, Q = {abc} \) . By some calculations, we get that\n\n\[ \mathrm{{LHS}} = \mathop{\sum }\limits_{{cyc}}\frac{1}{S + 1 - a} = \frac{{S}^{2} + {4S} + 3 + P}{{S}^{2} + {2S} + {PS} + P}, \]\n\n\[ \mathrm{{RHS}} = \mathop{\sum }\lim... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{abc}{2\left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) } \geq \frac{5}{3}. \] | Solution. WLOG, we may assume that \( a + b + c = 3 \) . Denote \( \mathrm{x} = {ab} + {bc} + {ca} \) and \( y = {abc} \) . Then we have\n\n\[ \frac{abc}{{a}^{3} + {b}^{3} + {c}^{3}} = \frac{y}{{27} + {3y} - {9x}} \]\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a}{b + c} = \frac{{27} + {3y} - {6x}}{{3x} - y}. \]\n\nWe need... | Yes |
Let \( a, b, c \) be real numbers with sum 3. Prove that\n\n\[ \left( {1 + a + {a}^{2}}\right) \left( {1 + b + {b}^{2}}\right) \left( {1 + c + {c}^{2}}\right) \geq 9\left( {{ab} + {bc} + {ca}}\right) . \] | Solution. We denote\n\n\[ x = a + b + c, y = {ab} + {bc} + {ca}, z = {abc}. \]\n\nAccording to the hypothesis, \( x = 3 \), so we can rewrite the inequality to\n\n\[ {z}^{2} + z + 1 + \mathop{\sum }\limits_{{sym}}(a + {a}^{2}) + \mathop{\sum }\limits_{{sym}}{ab} + \mathop{\sum }\limits_{{sym}}{a}^{2}{b}^{2} + {abc}\lef... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \frac{1}{1 + {3a}} + \frac{1}{1 + {3b}} + \frac{1}{1 + {3c}} + \frac{1}{1 + a + b + c} \geq 1. \] | Solution. We denote \( x = a + b + c \) and \( y = {ab} + {bc} + {ca} \) . Then the inequality can be rewritten to\n\n\[ \frac{3 + {6x} + {9y}}{{28} + {3x} + {9y}} + \frac{1}{1 + x} \geq 1 \Leftrightarrow \frac{1}{1 + x} \geq \frac{{25} - {3x}}{{28} + {3x} + {9y}} \]\n\n\[ \Leftrightarrow 3{x}^{2} - {19x} + {9y} + 3 \g... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 1. Prove that\n\n\[ \frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2} + {16abc}} \geq 8\left( {{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}}\right) . \] | Solution. Denote \( x = 4\left( {{ab} + {bc} + {ca}}\right) \) and \( y = {8abc} \) then we obtain\n\n\[ {a}^{2} + {b}^{2} + {c}^{2} = 1 - \frac{x}{2};\;{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2} = \frac{{x}^{2}}{16} - \frac{y}{4}. \]\n\nWe can rewrite the inequality to the form\n\n\[ {2x} \geq \left( {4 - {2x} +... | Yes |
Let \( a, b, c \) be non-negative real numbers with sum 2. Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} \geq 2\left( {{a}^{3}{b}^{3} + {b}^{3}{c}^{3} + {c}^{3}{a}^{3} + 4{a}^{2}{b}^{2}{c}^{2}}\right) .\n\]\n\n(Pham Kim Hung) | Solution. Write \( p = {ab} + {bc} + {ca} \) and \( q = {abc} \) . The inequality can be rewritten to\n\n\[ \n{\left( a + b + c\right) }^{2} - 2\left( {{ab} + {bc} + {ca}}\right) \geq 2{\left( ab + bc + ca\right) }^{3} - {6abc}\left( {a + b + c}\right) \left( {{ab} + {bc} + {ca}}\right) + {14}{a}^{2}{b}^{2}{c}^{2}.\n\]... | Yes |
Suppose that \( a, b, c \) are three real numbers satisfying \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove the following inequality\n\n\[ \n{a}^{3}\left( {b + c}\right) + {b}^{3}\left( {c + a}\right) + {c}^{3}\left( {a + b}\right) \leq 6.\n\] | Solution. Certainly, the inequality is non-homogeneous. However, the condition \( {a}^{2} + \) \( {b}^{2} + {c}^{2} = 3 \) can help change it to homogeneous as\n\n\[ \n{a}^{3}\left( {b + c}\right) + {b}^{3}\left( {c + a}\right) + {c}^{3}\left( {a + b}\right) \leq \frac{2}{3}{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \sqrt{\frac{{ab} + {bc} + {ca}}{3}} \leq \sqrt[3]{\frac{\left( {a + b}\right) \left( {b + c}\right) \left( {c + a}\right) }{8}}. \]\n | Solution. WLOG, suppose that \( {ab} + {bc} + {ca} = 3 \) . By AM-GM inequality, we deduce that \( a + b + c \geq 3 \) and \( {abc} \leq 1 \) . Therefore\n\n\[ \left( {a + b}\right) \left( {b + c}\right) \left( {c + a}\right) = \left( {a + b + c}\right) \left( {{ab} + {bc} + {ca}}\right) - {abc} = 3\left( {a + b + c}\r... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \n\frac{{\left( 2a + b + c\right) }^{2}}{2{a}^{2} + {\left( b + c\right) }^{2}} + \frac{{\left( 2b + c + a\right) }^{2}}{2{b}^{2} + {\left( c + a\right) }^{2}} + \frac{{\left( 2c + a + b\right) }^{2}}{2{c}^{2} + {\left( a + b\right) }^{2}} \leq 8.\n\]\n\n... | Solution. By normalizing the expression with \( a + b + c = 3 \), we reduce the left expression to a simpler form\n\n\[ \n\frac{{\left( 3 + a\right) }^{2}}{2{a}^{2} + {\left( 3 - a\right) }^{2}} + \frac{{\left( 3 + b\right) }^{2}}{2{b}^{2} + {\left( 3 - b\right) }^{2}} + \frac{{\left( 3 + c\right) }^{2}}{2{c}^{2} + {\l... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{{\left( b + c - a\right) }^{2}}{{\left( b + c\right) }^{2} + {a}^{2}} + \frac{{\left( c + a - b\right) }^{2}}{{\left( c + a\right) }^{2} + {b}^{2}} + \frac{{\left( a + b - c\right) }^{2}}{{\left( a + b\right) }^{2} + {c}^{2}} \geq \frac{3}{5}. \] | Solution. WLOG, we may suppose that \( a + b + c = 3 \) . The inequality becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{\left( 3 - 2a\right) }^{2}}{{a}^{2} + {\left( 3 - a\right) }^{2}} \geq \frac{3}{5} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{1}{2{a}^{2} - {6a} + 9} \leq \frac{3}{5}. \]\n\nJust notice tha... | Yes |
Let \( a, b, c \) be positive real numbers. Prove that\n\n\[ \n\\frac{{\\left( 2a + b + c\\right) }^{2}}{4{a}^{3} + {\\left( b + c\\right) }^{3}} + \\frac{{\\left( 2b + c + a\\right) }^{2}}{4{b}^{3} + {\\left( c + a\\right) }^{3}} + \\frac{{\\left( 2c + a + b\\right) }^{2}}{4{c}^{3} + {\\left( a + b\\right) }^{3}} \\le... | Solution. Suppose that \( a + b + c = 3 \) . The problem becomes\n\n\[ \n\\mathop{\\sum }\\limits_{{cyc}}\\frac{{\\left( 3 + a\\right) }^{2}}{4{a}^{3} + {\\left( 3 - a\\right) }^{3}} \\leq 4\n\]\n\nNotice that\n\n\[ \n\\frac{{\\left( 3 + a\\right) }^{2}}{4{a}^{3} + {\\left( 3 - a\\right) }^{3}} - \\frac{4}{3} = \\frac{... | Yes |
Let \( a, b, c, d \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{{b}^{2} + {c}^{2} + {d}^{2}} + \frac{b}{{c}^{2} + {d}^{2} + {a}^{2}} + \frac{c}{{d}^{2} + {a}^{2} + {b}^{2}} + \frac{d}{{a}^{2} + {b}^{2} + {c}^{2}} \geq \frac{3\sqrt{3}}{2} \cdot \frac{1}{\sqrt{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}}. \]\n | Solution. WLOG, assume that \( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = 1 \) . The problem becomes\n\n\[ \frac{a}{1 - {a}^{2}} + \frac{b}{1 - {b}^{2}} + \frac{c}{1 - {c}^{2}} + \frac{d}{1 - {d}^{2}} \geq \frac{3\sqrt{3}}{2}. \]\n\nBy AM-GM inequality, we get\n\n\[ 2{a}^{2}\left( {1 - {a}^{2}}\right) \left( {1 - {a}^{2}}... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{3\sqrt[3]{abc}}{2\left( {a + b + c}\right) } \geq 2. \] | Solution. Applying Cauchy-Schwarz inequality, we get\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a}{b + c} \geq \frac{{\left( a + b + c\right) }^{2}}{2\left( {{ab} + {bc} + {ca}}\right) }.\]\n\nWe normalize \( a + b + c = 1 \) and prove that\n\n\[ \frac{1}{2x} + \frac{3}{2}\sqrt[3]{abc} \geq 2 \]\n\nwhere \( x = {ab} + {b... | Yes |
Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ \frac{1}{{\left( a + b\right) }^{2}} + \frac{1}{{\left( b + c\right) }^{2}} + \frac{1}{{\left( c + a\right) }^{2}} \geq \frac{9}{4\left( {{ab} + {bc} + {ca}}\right) }.\] | Solution. We normalize \( {ab} + {bc} + {ca} = 1 \) . The inequality becomes\n\n\[ 4\mathop{\sum }\limits_{{cyc}}{\left( a + b\right) }^{2}{\left( a + c\right) }^{2} \geq 9{\left( a + b\right) }^{2}{\left( b + c\right) }^{2}{\left( c + a\right) }^{2} \]\n\nor\n\n\[ 4{\left( 1 + {a}^{2}\right) }^{2} + 4{\left( 1 + {b}^{... | Yes |
Let \( a, b, c, d \) be positive real numbers. Prove that\n\n\[ \frac{abc}{\left( {d + a}\right) \left( {d + b}\right) \left( {d + c}\right) } + \frac{bcd}{\left( {a + b}\right) \left( {a + c}\right) \left( {a + d}\right) } +\n\]\n\n\[ + \frac{cda}{\left( {b + a}\right) \left( {b + c}\right) \left( {b + d}\right) } + \... | Solution. Denote \( x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}, t = \frac{1}{d} \), the inequality becomes\n\n\[ \frac{{x}^{3}}{\left( {x + y}\right) \left( {x + z}\right) \left( {x + t}\right) } + \frac{{y}^{3}}{\left( {y + x}\right) \left( {y + z}\right) \left( {y + t}\right) } +\n\]\n\n\[ + \frac{{z}^{3}}{\le... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \frac{1}{3{a}^{2} + {\left( a - 1\right) }^{2}} + \frac{1}{3{b}^{2} + {\left( b - 1\right) }^{2}} + \frac{1}{3{c}^{2} + {\left( c - 1\right) }^{2}} \geq 1. \] | Solution. We want to find a real constant \( k \) such that\n\n\[ \frac{1}{3{a}^{2} + {\left( a - 1\right) }^{2}} \geq \frac{1}{3} + k\ln a. \]\n\nIf there exists such a valid number \( k \), we can conclude (notice \( \ln a + \ln b + \ln c = 0 \) )\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{1}{3{a}^{2} + {\left( a - 1\r... | Yes |
Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \frac{1}{{a}^{2} - a + 1} + \frac{1}{{b}^{2} - b + 1} + \frac{1}{{c}^{2} - c + 1} \leq 3. \] | Solution. First, we will prove that\n\n\[ f\left( x\right) = \frac{1}{{x}^{2} - x + 1} + \ln x - 1 \geq 0\forall x \in (0,{1.8}\rbrack . \]\n\nIndeed, we have\n\n\[ {f}^{\prime }\left( x\right) = \frac{-{2x} + 1}{{\left( {x}^{2} - x + 1\right) }^{2}} + \frac{1}{x} = \frac{\left( {x - 1}\right) \left( {{x}^{3} - {x}^{2}... | Yes |
Suppose that \( a, b, c, d, e, f \) are six positive real numbers satisfying abcdef \( = 1 \) . Prove the following inequality\n\n\[ \frac{{2a} + 1}{{a}^{2} + a + 1} + \frac{{2b} + 1}{{b}^{2} + b + 1} + \frac{{2c} + 1}{{c}^{2} + c + 1} + \frac{{2d} + 1}{{d}^{2} + d + 1} + \frac{{2e} + 1}{{e}^{2} + e + 1} + \frac{{2f} +... | Solution. Consider the following function with \( x > 0 \)\n\n\[ f\left( x\right) = \frac{1 + {2x}}{1 + x + {x}^{2}} + \frac{\ln x}{3} - 1. \]\n\nWe have certainly\n\n\[ {f}^{\prime }\left( x\right) = \frac{-2{x}^{2} - {2x} + 1}{{\left( 1 + x + {x}^{2}\right) }^{2}} + \frac{1}{3x} = \frac{\left( {x - 1}\right) \left( {... | Yes |
Suppose that \( a, b, c, d \) are positive real numbers satisfying abcd \( = 1 \) . Prove that\n\n\[ \frac{1 + a}{1 + {a}^{2}} + \frac{1 + b}{1 + {b}^{2}} + \frac{1 + c}{1 + {c}^{2}} + \frac{1 + d}{1 + {d}^{2}} \leq 4. \] | Solution. Consider the following function with \( x > 0 \)\n\n\[ f\left( x\right) = \frac{1 + x}{1 + {x}^{2}} + \frac{\ln x}{2} - 1 \]\n\nWe obtain\n\n\[ {f}^{\prime }\left( x\right) = \frac{\left( {x - 1}\right) \left( {{x}^{3} - {x}^{2} - {3x} - 1}\right) }{{2x}{\left( 1 + {x}^{2}\right) }^{2}}. \]\n\nSince the equat... | Yes |
Suppose that \( a, b, c, d, e, f \) are six positive real numbers satisfying abcdef \( = 1 \) . Prove the following inequality\n\n\[ \frac{a - 1}{{a}^{2} + a + 1} + \frac{b - 1}{{b}^{2} + b + 1} + \frac{c - 1}{{c}^{2} + c + 1} + \frac{d - 1}{{d}^{2} + d + 1} + \frac{e - 1}{{e}^{2} + e + 1} + \frac{f - 1}{{f}^{2} + f + ... | Solution. Consider the following function with \( x > 0 \)\n\n\[ f\left( x\right) = \frac{x - 1}{{x}^{2} + x + 1} - \frac{\ln x}{3}. \]\n\nNotice that\n\n\[ {f}^{\prime }\left( x\right) = \frac{\left( {x - 1}\right) \left( {-{x}^{3} - 6{x}^{2} - {3x} + 1}\right) }{{3x}{\left( {x}^{2} + x + 1\right) }^{2}}. \]\n\nThe eq... | Yes |
Suppose that \( a, b, c, d, e, f, g \) are positive real numbers such that \( a + b + c + d + e + f + g = 7 \) . Prove that\n\n\[ \left( {{a}^{2} - a + 1}\right) \left( {{b}^{2} - b + 1}\right) \left( {{c}^{2} - c + 1}\right) \left( {{d}^{2} - d + 1}\right) \left( {{e}^{2} - e + 1}\right) \left( {{f}^{2} - f + 1}\right... | Solution. The inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}\ln \left( {{a}^{2} - a + 1}\right) \geq 0 \]\n\nConsider the function \( f\left( x\right) = \ln \left( {{x}^{2} - x + 1}\right) - x + 1 \) . We have\n\n\[ {f}^{\prime }\left( x\right) = \frac{\left( {x - 1}\right) \left( {2 - x}\right) }{{x}^... | Yes |
Let \( a, b, c, d \) be positive real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} + \frac{1}{{c}^{2}} + \frac{1}{{d}^{2}} \geq {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}. \]\n\n(Pham Kim Hung) | Solution. Consider the following function of positive variable \( x \)\n\n\[ f\left( x\right) = \frac{1}{{x}^{2}} - {x}^{2} + {4x} - 4. \]\n\nWe clearly have\n\n\[ {f}^{\prime }\left( x\right) = \frac{-2}{{x}^{3}} - {2x} + 4 = \frac{-2\left( {x - 1}\right) \left( {{x}^{3} - {x}^{2} - x - 1}\right) }{{x}^{3}}. \]\n\nThe... | Yes |
Let \( a, b, c \) be non-negative real numbers such that \( a + b + c = 3 \) . Prove the following inequality\n\n\[ \left( {{a}^{2} - {ab} + {b}^{2}}\right) \left( {{b}^{2} - {bc} + {c}^{2}}\right) \left( {{c}^{2} - {ca} + {a}^{2}}\right) \leq {12}. \] | Solution. WLOG, assume that \( a \geq b \geq c \) . Certainly, we have\n\n\[ {b}^{2} - {bc} + {c}^{2} \leq {b}^{2} \]\n\n\[ {a}^{2} - {ac} + {c}^{2} \leq {c}^{2} \]\n\nIt suffices to prove that\n\n\[ M = {a}^{2}{b}^{2}\left( {{a}^{2} - {ab} + {b}^{2}}\right) \leq {12}. \]\n\nDenote \( x = \frac{a - b}{2} \geq 0 \) and ... | Yes |
Problem 2. Suppose that \( a, b, c \) are positive real numbers. Prove that\n\n\[ \left( {\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}}\\right) \\left( {\\frac{1}{1 + a} + \\frac{1}{1 + b} + \\frac{1}{1 + c}}\\right) \\geq \\frac{9}{1 + {abc}}. \]\n\n(Walther Janous) | Solution. By Rearrangement inequality,\n\n\[ \\frac{1}{a\\left( {1 + a}\\right) } + \\frac{1}{b\\left( {1 + b}\\right) } + \\frac{1}{c\\left( {1 + c}\\right) } \\geq \\frac{1}{b\\left( {1 + c}\\right) } + \\frac{1}{c\\left( {1 + a}\\right) } + \\frac{1}{a\\left( {1 + b}\\right) }\n\]\n\nHence the inequality will be pro... | Yes |
Problem 3. Let \( a, b, c \) be positive real numbers satisfying \( {a}^{2} + {b}^{2} + {c}^{2} = 3 \) . Prove that\n\n\[ \frac{a}{{a}^{2} + {2b} + 3} + \frac{b}{{b}^{2} + {2c} + 3} + \frac{c}{{c}^{2} + {2a} + 3} \leq \frac{1}{2}. \]\n\n(Pham Kim Hung) | Solution. We certainly have \( {a}^{2} + 1 \geq {2a},{b}^{2} + 1 \geq {2b},{c}^{2} + 1 \geq {2c} \), therefore\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a}{{a}^{2} + {2b} + 3} \geq \mathop{\sum }\limits_{{cyc}}\frac{a}{2\left( {a + b + 1}\right) } \]\n\nIt remains to prove that\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{a... | Yes |
Problem 4. Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive real numbers and \( {x}_{1}{x}_{2}\ldots {x}_{n} = 1 \) . Prove that\n\n\[ \n{x}_{1} + {x}_{2} + \ldots + {x}_{n} \geq \frac{2}{1 + {x}_{1}} + \frac{2}{1 + {x}_{2}} + \ldots + \frac{2}{1 + {x}_{n}}.\n\]\n\n(Pham Kim Hung) | Solution. Because \( 2 - \frac{2}{1 + {x}_{i}} = \frac{2{x}_{i}}{1 + {x}_{i}} \), the problem can be rewritten to\n\n\[ \n\mathop{\sum }\limits_{{cyc}}{x}_{i} + \mathop{\sum }\limits_{{cyc}}\frac{2{x}_{i}}{{x}_{i} + 1} \geq {2n}\n\]\n\nAccording to AM-GM inequality, we conclude that\n\n\[ \n- {2n} + \mathop{\sum }\limi... | Yes |
Prove that for all positive real numbers \( a, b, c \in \left\lbrack {1,2}\right\rbrack \), we have\n\n\[ \left( {a + b + c}\right) \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\right) \geq 6\left( {\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}}\right) . \] | Solution. Instead of the condition \( a, b, c \in \left\lbrack {1,2}\right\rbrack \), we will prove this inequality with a stronger condition that \( a, b, c \) are the side lengths of a triangle. Using the identities\n\n\[ \left( {\mathop{\sum }\limits_{{cyc}}a}\right) \left( {\mathop{\sum }\limits_{{cyc}}\frac{1}{a}}... | Yes |
Problem 6. Let \( a, b, c \) be positive real numbers such that \( a + b + c = 3 \) . Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} \geq \frac{2 + a}{2 + b} + \frac{2 + b}{2 + c} + \frac{2 + c}{2 + a}.\n\]\n(Pham Kim Hung) | Solution. (Cauchy reverse) By AM-GM, we deduce that\n\n\[ \n\mathop{\sum }\limits_{{cyc}}\frac{2 + a}{2 + b} = \mathop{\sum }\limits_{{cyc}}\frac{2 + a}{2} - \mathop{\sum }\limits_{{cyc}}\frac{b\left( {2 + a}\right) }{2\left( {2 + b}\right) } \geq \frac{9}{2} - \frac{3\sqrt[3]{abc}}{2}.\n\]\n\nSo it's enough to prove t... | Yes |
Problem 7. Let \( x, y, z \) be non-negative real numbers with sum 3. Prove that\n\n\[ \sqrt{\frac{x}{1 + {2yz}}} + \sqrt{\frac{y}{1 + {2zx}}} + \sqrt{\frac{z}{1 + {2xy}}} \geq \sqrt{3}. \] | Solution. According to Cauchy-Schwarz inequality, we deduce that\n\n\[ \mathop{\sum }\limits_{{cyc}}\sqrt{\frac{x}{1 + {2yz}}} = \mathop{\sum }\limits_{{cyc}}\frac{{x}^{2}}{\sqrt{x}\sqrt{{x}^{2} + 2{x}^{2}{yz}}} \]\n\n\[ \geq \frac{{\left( x + y + z\right) }^{2}}{\sqrt{x} \cdot \sqrt{{x}^{2} + 2{x}^{2}{yz}} + \sqrt{y} ... | Yes |
Problem 8. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be positive real numbers and \( {a}_{1}{a}_{2}\ldots {a}_{n} = 1 \) . Prove that\n\n\[ \sqrt{1 + {a}_{1}^{2}} + \sqrt{1 + {a}_{2}^{2}} + \ldots + \sqrt{1 + {a}_{n}^{2}} \leq \sqrt{2}\left( {{a}_{1} + {a}_{2} + \ldots + {a}_{n}}\right) \] | Solution. From the obvious inequality \( {\left( \sqrt{x} - 1\right) }^{4} \geq 0 \), we see that\n\n\[ \frac{1 + {x}^{2}}{2} \leq {\left( x - \sqrt{x} + 1\right) }^{2} \Rightarrow \sqrt{\frac{1 + {x}^{2}}{2}} + \sqrt{x} \leq 1 + x. \]\n\nAccording to this result, we have of course\n\n\[ \mathop{\sum }\limits_{{i = 1}}... | Yes |
Problem 9. Let \( a, b, c, k \) be positive real numbers. Prove that\n\n\[ \frac{a + {kb}}{a + {kc}} + \frac{b + {kc}}{b + {ka}} + \frac{c + {ka}}{c + {kb}} \leq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}. \]\n\n(Nguyen Viet Anh) | Solution. We denote\n\n\[ X = \frac{1 + k \cdot \frac{a}{b}}{1 + k}, Y = \frac{1 + k \cdot \frac{b}{c}}{1 + k}, Z = \frac{1 + k \cdot \frac{c}{a}}{1 + k}. \]\n\nAccording to Hölder inequality, we get\n\n\[ \mathop{\prod }\limits_{{cyc}}\left( {1 + \frac{ka}{b}}\right) \geq {\left( 1 + k\right) }^{3} \]\n\nor equivalent... | Yes |
Problem 10. Let \( a, b, c \) be positive real numbers such that \( {abc} = 1 \) . Prove that\n\n\[ \frac{1}{\left( {a + 1}\right) \left( {a + 2}\right) } + \frac{1}{\left( {b + 1}\right) \left( {b + 2}\right) } + \frac{1}{\left( {c + 1}\right) \left( {c + 2}\right) } \geq \frac{1}{2}. \] | Solution. By hypothesis \( {abc} = 1 \), so there are three positive real numbers \( x, y, z \) for which \( a = \frac{yz}{{x}^{2}}, b = \frac{zx}{{y}^{2}}, c = \frac{xy}{{z}^{2}} \) . The inequality becomes\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{{x}^{4}}{\left( {{x}^{2} + {yz}}\right) \left( {2{x}^{2} + {yz}}\right)... | Yes |
Problem 11. Let \( a, b, c, d \) be non-negative real numbers such that \( a + b + c + d = 4 \) . Prove that\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} - 4 \geq 4\left( {a - 1}\right) \left( {b - 1}\right) \left( {c - 1}\right) \left( {d - 1}\right) .\n\]\n\n(Pham Kim Hung) | Solution. Applying AM-GM inequality, we get\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} - 4 = {\left( a - 1\right) }^{2} + {\left( b - 1\right) }^{2} + {\left( c - 1\right) }^{2} + {\left( d - 1\right) }^{2}\n\]\n\n\[ \n\geq 4\sqrt{\left| {a - 1)\left( {b - 1}\right) \left( {c - 1}\right) \left( {d - 1}\right) }\righ... | Yes |
Problem 12. Let \( x, y, z \) be distinct real numbers. Prove that\n\n\[ \frac{{x}^{2}}{{\left( x - y\right) }^{2}} + \frac{{y}^{2}}{{\left( y - z\right) }^{2}} + \frac{{z}^{2}}{{\left( z - x\right) }^{2}} \geq 1 \] | Solution. We have\n\n\[ \mathop{\sum }\limits_{{cyc}}{\left( 1 - \frac{x}{z}\right) }^{2}{\left( 1 - \frac{z}{y}\right) }^{2} - {\left( 1 - \frac{y}{x}\right) }^{2}{\left( 1 - \frac{z}{y}\right) }^{2}{\left( 1 - \frac{x}{z}\right) }^{2} = \]\n\n\[ = \mathop{\sum }\limits_{{cyc}}{\left( 1 - \frac{x}{z} - \frac{z}{y} + \... | Yes |
Problem 13. Let \( a, b, c, d \) be non-negative real numbers with sum 3. Prove that\n\n\[ \n{ab}\left( {b + c}\right) + {bc}\left( {c + d}\right) + {cd}\left( {d + a}\right) + {da}\left( {a + b}\right) \leq 4.\n\] | Solution. WLOG, we may assume that \( b + d \leq a + c \) . We have\n\n\[ \n{ab}\left( {b + c}\right) + {bc}\left( {c + d}\right) + {cd}\left( {d + a}\right) + {da}\left( {a + b}\right) =\n\]\n\n\[ \n= \left( {a + c}\right) \left( {{bc} + {da}}\right) + \left( {b + d}\right) \left( {{ab} + {cd}}\right) =\n\]\n\n\[ \n= ... | Yes |
Problem 14. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be arbitrary real numbers. Prove that\n\n\[ \mathop{\sum }\limits_{{i, j = 1}}^{n}\left| {{a}_{i} + {a}_{j}}\right| \geq n\mathop{\sum }\limits_{{i = 1}}^{n}\left| {a}_{i}\right| \]\n\n(Iran TST 2006) | Solution. Separating the sequence \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) into two sub-sequences of nonnegative and elements\n\n\[ \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} = \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{r}}\right\} \cup \left\{ {-{c}_{1}, - {c}_{2},\ldots , - {c}_{s}}\right\} \]\n\nwith ... | Yes |
Problem 15. Let \( a, b, c \) be non-negative real numbers. Prove that\n\n\[ 3\left( {a + b + c}\right) \geq 2\left( {\sqrt{{a}^{2} + {bc}} + \sqrt{{b}^{2} + {ca}} + \sqrt{{c}^{2} + {ab}}}\right) . \] | Solution. WLOG, we may assume that \( a \geq b \geq c \) . Hence\n\n\[ \sqrt{{b}^{2} + {ca}} + \sqrt{{c}^{2} + {ab}} \leq \sqrt{2\left( {{b}^{2} + {c}^{2}}\right) + {2a}\left( {b + c}\right) }.\]\n\nWe need to prove that\n\n\[ 2\sqrt{2\left( {{b}^{2} + {c}^{2}}\right) + {2a}\left( {b + c}\right) } + 2\sqrt{{a}^{2} + {b... | Yes |
Problem 16. Suppose that \( a, b, c \) are three non-negative real numbers. Prove that\n\n\[ \frac{1}{{a}^{2} + {b}^{2}} + \frac{1}{{b}^{2} + {c}^{2}} + \frac{1}{{c}^{2} + {a}^{2}} \geq \frac{10}{{\left( a + b + c\right) }^{2}}. \]\n\n(Vasile Cirtoaje, Nguyen Viet Anh) | Solution. WLOG, assume that \( c = \min \left( {a, b, c}\right) \), then\n\n\[ {b}^{2} + {c}^{2} \leq {\left( b + \frac{c}{2}\right) }^{2} = {x}^{2} \]\n\n\[ {a}^{2} + {c}^{2} \leq {\left( a + \frac{c}{2}\right) }^{2} = {y}^{2} \]\n\n\[ {a}^{2} + {b}^{2} \leq {\left( a + \frac{c}{2}\right) }^{2} + {\left( b + \frac{c}{... | Yes |
Problem 17. Let \( a, b, c, d \) be positive real numbers such that \( {abcd} = 1 \) . Prove that\n\n\[ \left( {1 + {a}^{2}}\right) \left( {1 + {b}^{2}}\right) \left( {1 + {c}^{2}}\right) \left( {1 + {d}^{2}}\right) \geq {\left( a + b + c + d\right) }^{2}. \]\n\n(Pham Kim Hung) | Solution. Because \( {abcd} = 1 \), there are two numbers between \( a, b, c, d \) both not smaller than 1 or not bigger than 1. WLOG, suppose that they are \( b \) and \( d \), then \( \left( {b - 1}\right) \left( {d - 1}\right) \geq 0 \) . Applying Cauchy-Schwarz inequality, we get\n\n\[ \left( {1 + {a}^{2}}\right) \... | Yes |
Let \( a, b, c \) be positive real numbers such that \( a + b + c = 1 \) . Prove that\n\n\[ \frac{ab}{\sqrt{{ab} + {bc}}} + \frac{bc}{\sqrt{{bc} + {ca}}} + \frac{ca}{\sqrt{{ca} + {ab}}} \leq \frac{1}{\sqrt{2}}. \]\n\n(Chinese MO 2006) | Solution. The above inequality is equivalent to\n\n\[ \mathop{\sum }\limits_{{cyc}}a \cdot \sqrt{\frac{b}{a + c}} \leq \frac{1}{\sqrt{2}} \Leftrightarrow \mathop{\sum }\limits_{{cyc}}\frac{a + b}{2} \cdot \sqrt{\frac{{a}^{2}b}{\left( {a + c}\right) {\left( a + b\right) }^{2}}} \leq \frac{1}{\sqrt{2}}. \]\n\nUsing weigh... | Yes |
Let \( x, y, z \) be non-negative real numbers such that \( x + y + z = 1 \) . Find the maximum of\n\n\[ \frac{x - y}{\sqrt{x + y}} + \frac{y - z}{\sqrt{y + z}} + \frac{z - x}{\sqrt{z + x}} \] | Solution. First we consider the problem in the case \( \min \left( {x, y, z}\right) = 0 \) . WLOG, suppose that \( z = 0 \), then \( x + y = 1 \) and therefore\n\n\[ \mathop{\sum }\limits_{{cyc}}\frac{x - y}{\sqrt{x + y}} = \frac{x - y}{\sqrt{x + y}} + \sqrt{y} - \sqrt{x} = x - y + \sqrt{y} - \sqrt{x} = u\left( {v - 1}... | Yes |
Let \( x, y, z \) be three real numbers in \( \left\lbrack {-1,1}\right\rbrack \) such that \( x + y + z = 0 \) . Prove that\n\n\[ \sqrt{1 + x + {y}^{2}} + \sqrt{1 + y + {z}^{2}} + \sqrt{1 + z + {x}^{2}} \geq 3. \] | Solution. First, we will prove that if \( {ab} \geq 0 \) then\n\n\[ \sqrt{1 + a} + \sqrt{1 + b} \geq 1 + \sqrt{1 + a + b} \]\n\nIndeed, after squaring, the inequality becomes\n\n\[ 2 + a + b + 2\sqrt{\left( {1 + a}\right) \left( {1 + b}\right) } \geq 2 + a + b + 2\sqrt{1 + a + b} \]\n\n\[ \Leftrightarrow \left( {1 + a}... | Yes |
Suppose that \( a, b, c \) are three non-negative real numbers satisfying \( {ab} + {bc} + {ca} = 1 \) . Prove the inequality\n\n\[ \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \geq \frac{5}{2} \] | Solution. We denote \( x = a + b + c \) and \( z = {abc} \) . The inequality becomes\n\n\[ 2\mathop{\sum }\limits_{{cyc}}\left( {a + b}\right) \left( {a + c}\right) \geq 5\mathop{\prod }\limits_{{cyc}}\left( {a + b}\right) \Leftrightarrow 6 + 2\mathop{\sum }\limits_{{cyc}}{a}^{2} \geq 5\left( {a + b + c - {abc}}\right)... | Yes |
Prove the following inequality\n\n\\[ \n{\\left( \\sqrt{2}\\right) }^{n}\\left( {{a}_{1} + {a}_{2}}\\right) \\left( {{a}_{2} + {a}_{3}}\\right) ...\\left( {{a}_{n} + {a}_{1}}\\right) \\leq \\left( {{a}_{1} + {a}_{2} + {a}_{3}}\\right) \\left( {{a}_{2} + {a}_{3} + {a}_{4}}\\right) ....\\left( {{a}_{n} + {a}_{1} + {a}_{2... | Solution. According to the following results\n\n\\[ \n{\\left( {a}_{1} + {a}_{2} + {a}_{3}\\right) }^{2} \\geq \\left( {2{a}_{1} + {a}_{2}}\\right) \\left( {{a}_{2} + 2{a}_{3}}\\right) \n\\]\n\n\\[ \n\\left( {2{a}_{1} + {a}_{2}}\\right) \\left( {2{a}_{2} + {a}_{1}}\\right) = 2{a}_{1}^{2} + 2{a}_{2}^{2} + 5{a}_{1}{a}_{2... | Yes |
Problem 23. Let \( a, b, c \) be non-negative real numbers with sum 3. Prove that\n\n\[ \frac{{ab} + {bc} + {ca}}{{a}^{3}{b}^{3} + {b}^{3}{c}^{3} + {c}^{3}{a}^{3}} \geq \frac{{a}^{3} + {b}^{3} + {c}^{3}}{36}. \]\n\n(Pham Kim Hung, MYM) | Solution. WLOG, we may assume that \( a \geq b \geq c \) . Denote\n\n\[ f\left( {a, b, c}\right) = {36}\left( {{ab} + {bc} + {ca}}\right) - \left( {{a}^{3} + {b}^{3} + {c}^{3}}\right) \left( {{a}^{3}{b}^{3} + {b}^{3}{c}^{3} + {c}^{3}{a}^{3}}\right) .\n\nWe will prove \( f\left( {a, b, c}\right) \geq f\left( {a, b + c,0... | Yes |
Problem 24. Let \( a, b, c, d \) be four real numbers satisfying that \( \left( {1 + {a}^{2}}\right) \left( {1 + {b}^{2}}\right) (1 + \) \( \left. {c}^{2}\right) \left( {1 + {d}^{2}}\right) = {16} \) . Prove the inequality\n\n\[ - 3 \leq {ab} + {bc} + {ca} + {da} + {ac} + {bd} - {abcd} \leq 5. \] | Solution. We denote \( S = {ab} + {bc} + {cd} + {da} + {ac} + {bd} - {abcd} \), then\n\n\[ S - 1 = \left( {1 - {ab}}\right) \left( {{cd} - 1}\right) + \left( {a + b}\right) \left( {c + d}\right) . \]\n\nApplying Cauchy-Schwarz inequality, we obtain\n\n\[ {\left( S - 1\right) }^{2} \leq \left( {{\left( 1 - ab\right) }^{... | Yes |
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