Minimum value of $\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}$

Question

If $x,y,z>0$ and $x+y+z=60$. Then minimum value of $\displaystyle \frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}$

Try: $$\frac{(x+y)^2-2xy}{x+y}+\frac{(y+z)^2-2yz}{y+z}+\frac{(z+x)^2-2zx}{z+x}$$

So $$2(x+y+z)-2\bigg[\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}\bigg]$$

Now Using arithmetic geometric inequality

$x+y\geq 2\sqrt{xy}$ and $y+z\geq 2\sqrt{yz}$ and $z+x\geq 2\sqrt{zx}$

Could some help me how to solve further, Thanks

Answer 1

Use $$\frac{x^2+y^2}{x+y}\geq\frac{x+y}{2},$$ which is just C-S: $$2(x^2+y^2)=(1+1)(x^2+y^2)\geq(x+y)^2.$$ Indeed, $$\sum_{cyc}\frac{x^2+y^2}{x+y}\geq\sum_{cyc}\frac{x+y}{2}=60.$$ The equality occurs for $x=y=z=20,$ which says that we got a minimal value.

Answer 2

We have that

$$\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}=\sum_{cyc} \frac{(x+y)^2-2xy}{x+y}=120-\sum_{cyc} \frac{2xy}{x+y}\ge 120-3\cdot 20=60$$

indeed by AM-GM

$$\frac{x+y}2\ge \sqrt{xy}\implies \frac{2xy}{x+y}\le \sqrt{xy}$$

with equality for $x=y=z=20$.

Answer 3

By Engel's form of Cauchy Schwarz, $$\sum_{cyc}\frac{x^2+y^2}{x+y} = \sum_{cyc} \frac{x^2}{x+y} + \sum_{cyc} \frac{y^2}{x+y} \ge \frac{\left(\sum_{cyc} x\right)^2}{\sum_{cyc}(x+y)} + \frac{\left(\sum_{cyc} y\right)^2}{\sum_{cyc}(x+y)}\\ = \frac{\sum_{cyc} x}{2} + \frac{\sum_{cyc} y}{2} = \sum_{cyc} x = 60$$ Since the value $60$ is achieved at $x = y = z = 20$, $60$ is the desired minimum value.