Articles of cauchy schwarz inequality

A inequality proposed at Zhautykov Olympiad 2008

A inequality proposed at Zhautykov Olympiad 2008 Let be $a,b,c >0$ with $abc=1$. Prove that: $$\sum_{cyc}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$ $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. Our inequality becomes: $$\sum_{cyc}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$ Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{cyc}{\frac{z^2}{zx+y^2}} \geq \sum_{cyc}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$ Now applying Cauchy Schwarz we obtain the desired result . What I wrote can be found […]

If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$ After substitution $a=\frac{y}{x}$… I tried C-S, but without success.

Hard Olympiad Inequality

Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$. I tried to square expand homogenize then majorize. But I couldn’t make it work. Any help would be much appreciated.

If $a+b+c=6$ and $a,b,c$ belongs to positive reals $\mathbb{R}^+$; then find the minimum value of $\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$ .

If $a+b+c=6$ and $a,b,c$ belongs to positive reals, then find the minimum value of $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$$ using AM $\ge HM$ $\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ ${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{3}{2}$ or **why not $AM\ge GM $ $\frac{\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+a+b+c}{6}\ge (\frac{1}{a}\times\frac{4}{b}\times\frac{9}{c}\times a\times b\times c)^\frac{1}{6} $ $\Rightarrow \frac{1}{a}+\frac{4}{b}+\frac{9}{c}\ge 6(6^\frac{1}{3}-1)$**

$a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$

$a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$ Any ideas ?

Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$

Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$. I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further. Please help.

Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$

In the pdf which you can download here I found the following inequality which I can’t solve it. Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$ Thanks 🙂

Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$

Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$ Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$ My Try: By Cauchy Schwarz Inequality we have $$\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}$$ $\implies$ $$\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ Now by $AM \ge HM $ inequality we have $$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ $\implies$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}$$ hence $$\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}$$ hence $$3S \ge \frac{2601}{16}$$ so $$S \ge \frac{867}{16}$$ is […]

How prove this inequality $\sum\limits_{cyc}\frac{x^2}{(2x+y)(2x+z)}\le\frac{1}{3}$

let $x,y,z>0$,show that $$\dfrac{x^2}{(2x+y)(2x+z)}+\dfrac{y^2}{(2y+z)(2y+x)}+\dfrac{z^2}{(2z+x)(2z+y)}\le\dfrac{1}{3}$$ My try: $$\Longleftrightarrow\sum_{cyc}\dfrac{4x^2}{4x^2+2x(y+z)+yz}\le\dfrac{4}{3}$$ $$\Longleftrightarrow\sum_{cyc}\left(1-\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\right)\le\dfrac{4}{3} $$ $$\Longleftrightarrow \sum_{cyc}\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\ge\dfrac{5}{3}$$ then I can’t.Thank you

Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds. I spent a lot of time trying to solve this and, having consulted some books, I came to this: $$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 – x^2 […]