# Questions tagged [a.m.-g.m.-inequality]

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

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### $AM-GM$ for $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}$ [duplicate]

Let $a, b \in \mathbb{R_+}$ such that, $a + b = 1$. Show that $$(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}.$$ This was on a problem set for practising $AM-GM$ and I couldn’t ...
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### Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$

Given that $0 < a , b , c < 1$. Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$. I tried ...
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### finding x,y,z for optimizing expression

The question is as follows: let (x,y,z) be an ordered triplet of real numbers such that x<1 ,y<2 ,z<3 and x+ y/2 + z/3 >0. for x=a ,y=b,z=c the value of expression (1-x)(2-y)(3-z)(x+y/2+z/...
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### If $a+b+c+d=4$ Prove that $\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4$

Question - Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that $\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4$ My work ...
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### If $x+y+z=1$ prove $\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}$

Question - Let $x, y, z$ be non-negative real numbers with sum $1 .$ Prove that $$\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}$$ My work -...
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### $\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1$

Question Let.a, $b, c$ be positive real numbers with sum 3 . Prove that $$\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1$$ my doubt - by using cauchy reverse ...
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Suppose that $x,y$ are positive real numbers and that $$(x-y)^2=2\big( (x+y)-2\sqrt{xy} \big). \tag{*}$$ Then Mathematica claims that one of the following $3$ options holds: $$1. \, \, \, x=y.$$ $... 1answer 43 views ###$\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 $Question - Suppose that$a, b, c, d$are four positive real numbers with sum 4. Prove that $$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2$$ my doubt - ... 1answer 32 views ### Doubt in solution of problem in secrets in inequalities by pham kim hung Question - Suppose that$a, b, c$are three side-lengths of a triangle with perimeter 3. Prove that $$\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a} ... 0answers 19 views ### Proving |\prod_{i=0}^n(x-x_i)|\leq\frac{n!}{4}(\frac{b-a}{n})^{n+1} [duplicate] My task is to prove:$$ |\prod_{i=0}^n(x-x_i)|\leq\frac{n!}{4}(\frac{b-a}{n})^{n+1} $$where x_i=a+i\frac{b-a}{n} for i=0,...,n and x\in [a;b] I managed to prove that:$$ |(x-x_0)(x-x_1)|\leq \... 2answers 48 views ### Inequality$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$I found a problem where now I essentially have to show the following inequality $$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$$ where$a,b,c$are the sides of a triangle. I have tried a lot of ... 1answer 41 views ### Doubt in solution of APMO 1998 Inequality problem Question - Let$a, b, c$be positive real numbers. Prove that $$\begin{array}{c} \left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt{x y z}}... 2answers 45 views ### Minimize surface of cone with given volume without using derivatives Given volume of cone equal to V, i need to minimize side surface area of given cone without using derivatives. Exactly i'm trying to use AM-GM inequality. First of all i tried to do it like it was ... 1answer 66 views ### linear programming with logarithm max log x_1 + a log x_2 subject to 2$$x_1$+ $x_2$ $\le 8$ $x_1$+$x_2$ $\le 5$ $x_1 \ge 0$ , $x_2 \ge 0$ i need to solve and show optimal solution when a =1 but how can i find it? is it ...
If $x,y,z>0$ and $x(1-y)>{1\over 4}$ , $y(1-z)>{1\over 4}$ , $z(1-x)>{1\over 4}$ then find the number of order triplets (x,y,z) satisfying the above inequaltiy. I am stuck as I know how ...
### Does $|f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|$ imply $f$ is a logarithm?
Let $f:\mathbb R^+ \to \mathbb R$ be a continuous function, and suppose that $$|f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|, \tag{1}$$ holds for every $x,y \in \mathbb R^+$. Suppose also that \$f(1)=...