$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac{1}{2}(x+y+z)$

Question

let $x,y,z$ be positive real numbers $$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac{1}{2}(x+y+z)$$

then how to prove above by Cauchy Schwarz -Inquality

Answer 1

The standard fact that $a^{2}+b^{2}\geq 2ab$ for $a,b\geq 0$ follows by C-S: Consider $(a,b)$ and $(b,a)$, then $2ab=(a,b)\cdot(b,a)\leq\|(a,b)\|\cdot\|(b,a)\|=a^{2}+b^{2}$.

Now we have $\dfrac{a^{2}}{b}\geq 2a-b$ and we put $a=2x$ and $b=y+z$ we get \begin{align*} \dfrac{(2x)^{2}}{y+z}\geq 4x-y-z. \end{align*} Doing this in the symmetric fashion we get \begin{align*} \dfrac{(2x)^{2}}{y+z}+\dfrac{(2y)^{2}}{x+z}+\dfrac{(2z)^{2}}{x+y}\geq 4(x+y+z)-2(x+y+z)=2(x+y+z). \end{align*}

Answer 2

By C-S: $$\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}.$$

Answer 3

Well now I got it.

Consider the two sets of real numbers $${\frac {x}{\sqrt {y+z}}, \frac {y}{\sqrt {x+z}}, \frac {z}{\sqrt {y+x}}}$$

and $${\sqrt {y+z}, \sqrt {x+z},\sqrt {y+x}}$$ Now using CS we get $$\left(\frac {x^2}{y+z}+ \frac {y^2}{x+z}+ \frac {z^2}{y+x}\right)\left(2(x+y+z)\right) \ge {(x+y+z)}^2$$

Hence we get

$$\left(\frac {x^2}{y+z}+ \frac {y^2}{x+z}+ \frac {z^2}{y+x}\right) \ge \frac{x+y+z}{2}$$

Q. E. D

Answer 4

Proof 1: Another quite well known proof to this classic problem is as follows: WLOG assume $x+y+z = 1$ and this is possible since you can rewrite the left side of the inequality as $\displaystyle \sum_{\text{cyclic}}\dfrac{\left(\dfrac{x}{x+y+z}\right)^2}{\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}}\ge \dfrac{1}{2}$. With the assumption $\displaystyle \sum_{\text{cyclic}} x = 1$ in mind, the inequality becomes: $\displaystyle \sum_{\text{cyclic}} \dfrac{x^2}{1-x} \ge \dfrac{1}{2}$. To this end, consider $f(x) = \dfrac{x^2}{1-x} = -1-x-\dfrac{1}{x-1}$, and $f''(x)=\dfrac{2}{(1-x)^3}>0$. Thus $f$ is convex, and $LHS = \displaystyle \sum_{\text{cyclic}} f(x)\ge 3f\left(\dfrac{x+y+z}{3}\right)=3f\left(\dfrac{1}{3}\right)=3\times \dfrac{1}{6} = \dfrac{1}{2}$ as claimed. Note: the word "cyclic" can be removed, but I still put it there because majority of problems similar to this one can be written nicely with the short cut using the word "cyclic".

Proof 2: This is probably what OP is craving for...since it is simple and easy to see...To this end again, use the same assumption in proof 1 above and the function $f(x) = -1-x +\dfrac{1}{1-x}$. Thus using AM-GM inequality: $LHS = f(x)+f(y)+f(z) = (-1-x+\dfrac{1}{1-x})+(-1-y+\dfrac{1}{1-y})+(-1-z+\dfrac{1}{1-z})= \dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}-4\ge \dfrac{9}{(x+y)+(y+z)+(z+x)}-4=\dfrac{9}{2}-4 = \dfrac{1}{2}$. Note: This can be done by using the Angel Form of the CS inequality, but we view it as a AM-GM inequality for we applied it twice.

Answer 5

Also, SOS helps: $$\sum_{cyc}\left(\frac{x^2}{y+z}-\frac{x}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{x(x-y-(z-x))}{y+z}=$$ $$=\frac{1}{2}\sum_{cyc}(x-y)\left(\frac{x}{y+z}-\frac{y}{x+z}\right)=\frac{1}{2}\sum_{cyc}\frac{(x-y)^2(x+y+z)}{(x+z)(y+z)}\geq0.$$

Solution by Tangent Line method.

Since our inequality is homogeneous, we can assume $x+y+z=3$.

Thus, we need to prove that $$\sum_{cyc}\frac{x^2}{3-x}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(\frac{x^2}{3-x}-\frac{1}{2}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)(2x+3)}{3-x}\geq0$$ or $$\sum_{cyc}\left(\frac{(x-1)(2x+3)}{3-x}-\frac{5}{2}(x-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)^2}{3-x}\geq0.$$