Articles of cauchy schwarz inequality

Solve this with CBS: minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$

How can you see the minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ using the CBS inequality? I have seen a proof of that that use trigonometric substitutions, but I don’t see as one-step the solution using the CBS inequality.

About the Application of Cauchy-Schwarz (Basic): maximum of $3x+4y$ for $x^2+y^2 \leq 16$

Today I’ve seen a question like this: $$\text{Given } x^2+y^2 \leq 16, \text{ what is the maximum value for } 3x+4y?$$. What I’ve tried was the following: $$3x+4y \leq \sqrt{x^2+y^2}\cdot\sqrt{9+16}$$ But the problem here is that I have to give a value for $\sqrt{x^2 +y^2}$ and I know that this would not give me a […]

How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$?

How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$ for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$. I do not know where to start. I need some idea and advice on this problem.Thanks

If $xy+xz+yz=3$ so $\sum\limits_{cyc}\left(x^2y+x^2z+2\sqrt{xyz(x^3+3x)}\right)\geq2xyz\sum\limits_{cyc}(x^2+2)$

Prove that for any set of three positive real $x, y, z$ such that $xy+yz+zx=3$ $x^2(y+z)+y^2(x+z)+z^2(x+y)+2\sqrt {xyz}\left(\sqrt{x^3+3x}+\sqrt{y^3+3x}+\sqrt{z^3+3x}\right)\ge$ $\ge 2xyz(x^2+y^2+z^2+6)$ Reasoning I know that the minimum value is with $x=y=z$ We have to find an equation equivalent to the date where we subtract and divide by xyz and don’t appear other variables.

Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$

For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$ We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{\left(z+1\right)^2\left(z^2+1\right)}+\frac{1}{\left(y+1\right)^2\left(y^2+1\right)}+\frac{1}{\left(x+1\right)^2\left(x^2+1\right)}\right)$ $\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{2\left(z^2+1\right)^2}+\frac{1}{2\left(x^2+1\right)^2}+\frac{1}{2\left(y^2+1\right)^2}\right)$ $\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{9}{2\left(\left(z^2+1\right)^2+\left(y^2+1\right)^2+\left(x^2+1\right)^2\right)}\right)$ I can’t continue. Help

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 ?