Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$

Question

$x,y,z \in \mathbb R^+$ different than $0$, Prove that : $$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

My attempt: First, we should prove that : $$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$ I will consider it as proven, we gonna start by : $$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$

by the same method we have : $x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$

so we can replace $2xy$ by $x^2+y^2$ in this inequality $$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$ so :

$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

I'm new to AM-GM inequality, if it exists a better proof using it please post it.

Answer 1

Probably the following is shorter and easier.

Firstly, note that for positive reals $a$ and $b$ the inequality $$ \frac{a^2}{b}\ge 2a-b $$ holds (since it's equivalent to $\frac{(a-b)^2}{b}\ge 0$). Applying this observation six times gives $$ \frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\ge(2x-z+2y-z)+(2x-y+2z-y)+(2y-x+2z-x)= \\ =2(x+y+z), $$ as desired.

Comment. One can generalize this trick, namely the inequality $$ \frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y} $$ holds for $a,b,x,y>0$, $x\neq y$. This is often helpful when we need to estimate some fractions using polynomial expressions.

Answer 2

$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$ $$\Leftrightarrow (x^2 + y^2 + z^2) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3(x+y+z)$$ Note that $$(x^2 + y^2 + z^2) \geq \frac{1}{3} (x+y+z)^2 $$ (Cauchy-Schwarz ineq., equality: x = y = z) and $$(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq (1+1+1)^2 = 9$$ (Cauchy-Schwarz ineq., equality: $x=y=z$).

Applying them implies the original inequality (equality holds when $x=y=z$).