Inequalities

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sqing

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sqing

#1 h Oct 31, 2024, 2:22 PM

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Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that
$2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})}$ $2a+9b\geq 6\sqrt[4]{48\sqrt{2}-39}$

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#2 h Nov 1, 2024, 3:27 AM

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Let $a,b>0$ and $a+2b\leq 8.$ Prove that
$ab^2(2a+b)\leq \frac{471+133\sqrt{57}}{9}$ Let $a,b>0$ and $a+3b\leq 4.$ Prove that
$ab^2(a+b)\leq \frac{39+55\sqrt{33} }{144}$

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#3 h Nov 1, 2024, 7:54 AM • 1 Y

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sqing wrote:

Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that $2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})}$

solution

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#4 h Nov 1, 2024, 8:18 AM

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Very very nice.Thank DAVROS.
Let $a,b >0 .$ Prove that
$\frac{a^2}{b+1}+ \frac{b^2}{a+1}\geq \frac{a^2+b^2}{\sqrt{ab}+1}$ Let $a,b,c >0 .$ Prove that $\frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}\geq \frac{3abc}{abc+1}$ $\frac{a}{(a+b)^2+c^2}+ \frac{b}{(b+c)^2+a^2}+ \frac{c}{(c+a)^2+b^2}\leq \frac{9}{5(a+b+c)}$

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#5 h Nov 1, 2024, 9:10 AM

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Let $a,b>0$ and $a+b=2.$ Prove that
$\left(\frac{311}{50}ab +1 \right)\left(a+\frac{1}{b} +\frac{b}{a}\right) \geq \frac{1083}{50}$ $\left(\frac{1247}{200}ab +1 \right)\left(a+\frac{1}{b} +\frac{b}{a}\right) \geq \frac{4341}{200}$

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#6 h Nov 1, 2024, 10:44 AM

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sqing wrote:

Let $a,b>0$ and $a+2b\leq 8.$ Prove that $ab^2(2a+b)\leq \frac{471+133\sqrt{57}}{9}$

solution

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#7 h Nov 1, 2024, 11:42 AM

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Very very nice.Thank DAVROS.
Let $a,b,c\geq 0 .$ Prove that $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4c}{a+b+c}\geq 2$ $\sqrt{\frac{a}{b+c}}+ \sqrt{\frac{b }{c+a}}+ 2\sqrt{\frac{c}{a+b+c}}\geq 2$ Let $a,b>0 .$ Prove that $\frac{a^3}{b(a+1)^2}+ \frac{b^3}{a(b+1)^2}\geq \frac{ab}{ab+1}$

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#8 h Nov 1, 2024, 12:43 PM

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Let $a,b>0$ and $(a^{12}+1)(b^{12}+1)\le 4 .$ Prove that $a+b+ ab \leq 3$

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#9 h Nov 1, 2024, 6:46 PM

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I'm not very good at inequalities so idk if my solution works but here it is anyways:

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#10 h Nov 2, 2024, 2:10 AM

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Thanks.
Let $a,b,c\geq 0 .$ Prove that $\frac{a}{(b+c)^2}+ \frac{b }{(c+a)^2}+ \frac{4}{a+b}\geq \frac{2(a+b+c)}{ab+bc+ca}$ Let $a,b,c>0$ and $a+b+ab\geq 3.$ Prove that $\frac{a}{b}+ \frac{1}{a}+b\geq \frac{9}{2ab+1}$ Let $a,b,c>0$ and $ab+bc+ca+abc\geq 4.$ Prove that $\frac{a}{b}+ \frac{c}{a}+ \frac{1}{c}+b\geq \frac{16}{3abc+1}$

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#11 h Nov 2, 2024, 2:54 AM

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sqing wrote:

Let $a,b>0$ and $(a^{12}+1)(b^{12}+1)\le 4 .$ Prove that $a+b+ ab \leq 3$

Solution. From the assumption and AM-GM inequality, it follows that
$\begin{align*}36\ge&32+ (a^{12}+1)(b^{12}+1)\\ =& 11+a^{12}+11+b^{12}+11+(ab)^{12}\\ \ge&12a+12b+12ab, \end{align*}$ thereby $a+b+ab\le3$ , and the maximum value $3$ can be attained if $a=b=1$ . $\blacksquare$

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#12 h Nov 2, 2024, 3:37 AM

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Very very nice.Thank ytChen.
Let $a,b,c\geq 0 .$ Prove that $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4(c+\sqrt{ab})}{a+b}\geq 4$ $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{9c+\sqrt{ab}}{a+b}\geq 6$ $\frac{a}{b+c}+ \frac{b+\sqrt{ca}}{c+a}+ \frac{c+\sqrt{ab}}{a+b}\geq 2$ $\frac{a+\sqrt{bc}}{b+c}+ \frac{b+\sqrt{ca} }{c+a}+ \frac{ c+\sqrt{ab}}{a+b} \geq \frac52$

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#13 h Nov 2, 2024, 4:53 AM

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Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that
$2\leq \frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 } \leq \frac{1000}{243}$ $\frac{1000}{1701} \leq\frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 +2}\leq \frac{2}{3}$

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#14 h Nov 2, 2024, 10:45 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that $2\leq \frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 } \leq \frac{1000}{243}$

solution

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#15 h Nov 3, 2024, 9:32 AM

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Very very nice.Thank DAVROS.
Let $a,b> 0$ and $a+b=2.$ Prove that
$(a +\sqrt{a^2 +3b}) (b +\sqrt{b^2 +3a})\geq 9$ $\frac{a}{b+\sqrt{a^2 +3b}}+\frac{b}{a+\sqrt{b^2 +3a}}\geq \frac{2}{3}$ $\frac{1}{a+\sqrt{b^2 +3a}}+\frac{1}{b+\sqrt{a^2 +3b}}\geq \frac{2}{3}$ https://artofproblemsolving.com/community/c6h2338297p18827338

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#46 h Nov 17, 2024, 1:15 AM

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Very very nice .Thank ytChen .
Let $a,b,c,d$ be reals such that $a^2+b^2+c^2+d^2=1.$ Prove that
$(a+b+c )d\leq \frac{\sqrt 3}{2}$ Let $a,b,c,d > 0$ and $a^3+b^3+c^3+d^3=1.$ Prove that
$(a^2+b^2+c^2)d \leq \sqrt[3]{\frac{4}{9}}$ Let $a,b,c$ be reals . Prove that
$\frac{a+b}{(16a^4+7)(16b^4+7)} \leq \frac{1}{64}$ $\frac{a+b+c}{(9 a^2+5)(9 b^2+5)(9c^2+5)} \leq \frac{1}{216}$ Let $a,b,c > 0 .$ Prove that
$\frac{a+b}{(8a^3 +5)(8b^3+5)} \leq \frac{1}{36}$ $\frac{a+b}{(32a^5+9)(32b^5+9)} \leq \frac{1}{100}$ $\frac{a+b+c}{(27a^3+8)(27 b^3+8)(27c^3+8)} \leq \frac{1}{729}$ Let $a,b,c,d$ be reals . Prove that
$\frac{a+b+c+d}{(16a^2+7)(16b^2+7)(16c^2+7)(16d^2+7)} \leq \frac{1}{4096}$

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#47 h Nov 20, 2024, 11:33 PM

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sqing wrote:

Let $a,b,c >0 .$ Prove that $\frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}\geq \frac{3abc}{abc+1}$

Solution. First recall inequality $(x+y+z)^2\ge3(xy+yz+zy).$ Thus, by Cauchy-Schwartz’s inequality, we get
$\begin{align*}& \frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}=\\ & \frac{a^6}{a^3\left(b^3+1\right)}+ \frac{b^6}{b^3\left(c^3+1\right)}+ \frac{c^6}{c^3\left(a^3+1\right)}\\ \ge& \frac{\left(a^3+b^3+c^3\right)^2}{a^3b^3+b^3c^3+c^3a^3+ a^3+b^3+c^3}\\ =& \frac{1}{\frac{a^3b^3+b^3c^3+ c^3a^3}{\left(a^3+b^3+c^3\right)^2 }+\frac{1}{a^3+b^3+c^3}}\\ \ge&\frac{1}{\frac{1}{3}+\frac{1}{3abc}}=\frac{3abc}{abc+1}, \end{align*}$ and the equality occurs if $a=b=c=1$ . $\blacksquare$

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#48 h Nov 21, 2024, 1:12 AM

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Very very nice .Thank ytChen .
Let $a,b,c\ge 0$ and $ab+bc+ca+abc=4$ .Prove that
$(a+b+c-4)^2+4 \ge 5abc$ Equality holds when $(a,b,c)=\left(\frac{2}{3}, \frac{2}{3},2\right).$

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#49 h Nov 21, 2024, 1:42 AM

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sqing wrote:

Let $a,b,c >0 .$ Prove that
$\frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2} \leq \frac{3}{5}$

Solution.Cauchy-Schwartz’s inequality shows
$\begin{align*}& \frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2}\\ \le& \frac{ab}{4ab+c^2}+\frac{bc}{4bc+a^2}+ \frac{ca}{4ca+b^2}\\ =& \frac{1}{4}\left[3-\frac{c^2}{4ab+c^2}-\frac{a^2}{4bc+a^2}- \frac{b^2}{4ca+b^2}\right]\\ \le& \frac{1}{4}\left[3-\frac{(a+b+c)^2}{(a+b+c)^2+2(ab+bc+ac)}\right]\\ =& \frac{1}{4}\left[3-\frac{1}{1+\frac{2(ab+bc+ca)}{(a+b+c)^2}}\right]\le \frac{1}{4}\left[3-\frac{1}{1+\frac{2}{3}}\right] =\frac{3}{5}, \end{align*}$ and the maximum value $\frac{3}{5}$ can be attained if $a=b=c$ . $\blacksquare$

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#50 h Nov 21, 2024, 1:51 AM

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Very very nice .Thank ytChen .
Let $a,b,c>0.$ Prove that
$\frac{\frac{a^2}{b+c}}{\frac{b}{c+a}+\frac{c}{a+b}}+\frac{\frac{b^2}{c+a}}{\frac{c}{a+b}+\frac{a}{b+c}} +\frac{\frac{c^2}{a+b}}{\frac{a}{b+c}+\frac{b}{c+a}} \geq\frac{a+b+c}{2}$ $\frac{1}{\frac{a}{a+b}+\frac{b}{b+c}} +\frac{1}{\frac{b}{b+c}+\frac{c}{c+a}} +\frac{1}{\frac{c}{c+a}+\frac{a}{a+b}} >\frac{5}{2}$ $\frac{(a^2+bc)(b^2+ca)(c^2+ab)}{abc}\geq \frac{1}{2}(a+b+c)^3$ Let $a,b \in [0,1] .$ Prove that
$\frac{1}{1+a+b+c+abc}\geq 1-\frac{a+b+c}{3}+\frac{abc}{5} \geq \frac{1}{2+ab+bc+ca}$ Let $a,b,c\geq 0,a+b+c=4$ and $a^2+b^2+c^2=10 .$ Prove that $a^3b+b^3c+c^3a+8abc\leq 27$ Let $a,b,c>0$ and $a+b+c=3 .$ Prove that
$\frac{a+b}{c+a+1}+\frac{b+c}{a+b+1}+\frac{c+a}{b+c+1} \geq 2 \geq \frac{a(b+c)}{a+b+1}+\frac{b(c+a)}{b+c+1}+\frac{c(a+b)}{c+a+1}$ Let $a,b,c>0$ and $abc=1 .$ Prove that
$\frac{a+b}{c+a+1}+\frac{b+c}{a+b+1}+\frac{c+a}{b+c+1} \geq 2$ Let $a,b,c>0$ and $a^2+b^2+c^2=3 .$ Prove that
$\frac{a(b+c)}{a+b+1}+\frac{b(c+a)}{b+c+1}+\frac{c(a+b)}{c+a+1} \leq 2$

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#51 h Nov 21, 2024, 2:49 AM

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Let $a,b,c\geq 0 .$ Prove that $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{ c+\sqrt{ab} }{a+b}\geq 2$

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#52 h Nov 21, 2024, 7:46 AM

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sqing wrote:

Let $a,b,c\geq 0 .$ Prove that $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{ c+\sqrt{ab} }{a+b}\geq 2$

Solution. Without loss of generality, assume $a\ge b\ge0$ . For $\frac{a+b }{c+\sqrt{ab}}+ \frac{ c+\sqrt{ab} }{a+b}\ge2$ , it suffices to show
$\frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\ge0.$ Indeed,
$\begin{align*}& \frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\\ =& \frac{a}{b+c}-\frac{a}{c+\sqrt{ab} } + \frac{b }{c+a}-\frac{b}{c+\sqrt{ab} }\\ =& \frac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{(b+c)\left(c+\sqrt{ab}\right)}+ \frac{b\sqrt a\left(\sqrt b-\sqrt a\right)}{(c+a)\left(c+\sqrt{ab}\right)}\\ \ge& \frac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{(a+c)\left(c+\sqrt{ab}\right)}+ \frac{b\sqrt a\left(\sqrt b-\sqrt a\right)}{(c+a)\left(c+\sqrt{ab}\right)}\\ =& \frac{\sqrt{ab}\left(\sqrt a-\sqrt b\right)^2}{(a+c)\left(c+\sqrt{ab}\right)}\ge0, \end{align*}$ and the equality occurs if $b=0$ and $a=c$ . $\blacksquare$

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#53 h Nov 21, 2024, 8:09 AM

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Very very nice.Thank ytChen:
Let $a,b,c > 0 .$ Then $\frac{a}{b+c}+ \frac{b }{a+c} \geq\frac{a+b}{\sqrt{ab}+c}$ h

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#54 h Nov 21, 2024, 8:31 AM

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Let $a,b > 0 .$ Prove that $\frac{a^2}{b+1}+ \frac{b^2 }{a+1} \geq\frac{a^2+b^2}{\sqrt{ab} +1}$

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#55 h Nov 21, 2024, 8:42 AM

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sqing wrote:

Let $a,b,c\geq 0 .$ Prove that $\frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4(c+\sqrt{ab})}{a+b}\geq 4$

Solution. Without loss of generality, assume $a\ge b\ge0$ . Since $\frac{a+b }{c+\sqrt{ab}}+ \frac{4\left(c+\sqrt{ab}\right)}{a+b}\ge4,$ it suffices to show
$\frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\ge0,$ which was proved in post #53, and the equality occurs if $b=0$ and $a=2c$ . $\blacksquare$

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#56 h Nov 21, 2024, 9:06 AM

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Very very nice.Thank ytChen.
Let $a,b,c>0$ and $a+b+c=3 .$ Prove that
$\frac{1}{a^3+abc+1}+\frac{1}{b^3+abc+1}+\frac{1}{c^3+abc+1} \geq 1$ $\frac{a(a+b)}{c(c+1)}+\frac{b(b+c)}{a(a+1)}+\frac{c(c+a)}{b(b+1)} \geq 3$ $\frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1} \geq 3 \geq \frac{a(b+c)}{a+1}+\frac{b(c+a)}{b+1}+\frac{c(a+b)}{c+1}$ $\frac{(a+b)(b+c)(c+a)}{(a+2)(b+2)(c+2)} \leq \frac{8}{27}$ https://artofproblemsolving.com/community/c6h1338925p7259810
Let $a,b,c>0$ and $a+b+c =1.$ Prove that
$\frac{a+b}{a+bc}+\frac{b+c}{b+ca}+\frac{c+a}{c+ab} \geq \frac{9}{2}$

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#57 h Nov 27, 2024, 3:57 AM

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sqing wrote:

Let $a,b>0 .$ Prove that $\frac{a^3}{b(a+1)^2}+ \frac{b^3}{a(b+1)^2}\geq \frac{ab}{ab+1}$

Solution. Using Hölder’s inequality and AM-GM inequality, we get
$\begin{align*} &\frac{a^3}{b(a+1)^2}+ \frac{b^3}{a(b+1)^2}\\ \ge& \frac{(a+b)^3}{2\left[b(a+1)^2+a(b+1)^2\right]}\\ =& \frac{(a+b)^3}{2\left[ab(a+b)+4ab+(a+b)\right]}\\ =& \frac{1}{2\left[\frac{ab}{(a+b)^2}+\frac{4ab}{(a+b)^3}+\frac{1}{(a+b)^2}\right]}\\ \ge& \frac{1}{2\left[\frac{ab}{4ab}+\frac{4ab}{8ab\sqrt{ab}}+\frac{1}{4ab}\right]}\\ =&\frac{1}{\frac{1}{2}\left[1+\frac{2}{\sqrt{ab}}+\frac{1}{ab}\right]}\\ =& \frac{ab}{\frac{1}{2}\left(\sqrt{ab}+1\right)^2}\ge\frac{ab}{ab+1}, \end{align*}$ where the equality occurs if $a=b=1$ . $\blacksquare$

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#58 h Nov 27, 2024, 6:01 AM

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Very very nice.Thank ytChen.
Let $a,b,c>0 .$ Prove that
$\frac{a^2(a+b)}{b(b+c)}+\frac{b^2(b+c)}{c(c+a)}+\frac{c^2(c+a)}{a(a+b)} \geq a+b+c$ $\frac{a^2(a+b)}{b(b+c)}+\frac{b^2(b+c)}{c(c+a)}+\frac{c^2(c+a)}{a(a+b)} \geq \sqrt{3(a^2+b^2+c^2)}$

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#59 h Dec 22, 2024, 5:22 AM

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sqing wrote:

Let $x\geq 0 .$ Prove that
$\frac{(\sqrt{4+9x}+\sqrt{ x })\sqrt{1+2x}}{4+9x}\geq \frac 12$

sqing wrote:

Let $x\geq 0 .$ Prove that
$\frac{(\sqrt{4+9x}-\sqrt{ x })\sqrt{1+2x}}{4+9x}\leq \frac 12$

Solution.

Since $x\ge0$ , we get that
$\begin{align*}&\frac{(\sqrt{4+9x}-\sqrt{ x })\sqrt{1+2x}}{4+9x}\\ \leq& \frac{\sqrt{4+9x}\cdot\sqrt{1+2x}}{4+9x}\\ =& \sqrt{\frac{1+2x}{4+9x}}\le\sqrt{\frac14}=\frac 12, \end{align*}$ the maximum value $\frac{1}{2}$ can be attained if $x=0$ .
From $x\ge0$ , it follows that
$\begin{align*}& 2\cdot\frac{(\sqrt{4+9x}+\sqrt{ x })\sqrt{1+2x}}{4+9x}\\ \ge& 2\cdot\frac{(\sqrt{4+8x}+\sqrt{ x })\sqrt{1+2x}}{4+9x}\\ =& 2\cdot\frac{2(1+2x)+\sqrt{ x }\sqrt{1+2x}}{4+9x}\\ \ge& 2\cdot\frac{2(1+2x)+x}{4+9x}\\ =&\frac{4+10x}{4+9x}\ge1, \end{align*}$ hence the result, and the minimum value $\frac{1}{2}$ can be attained if $x=0$ . $\blacksquare$

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#60 h Dec 22, 2024, 7:30 AM

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Very very nice.Thank ytChen.

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