lagrange multiplier inequality question

Question

Prove for every $x,y,z>0, that\ f(x,y,z)= \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\geq \frac{3}{2} $

So since $f$ is a homogeneous function we can prove this on the following constraint $g(x,y,z)=x+y+z-1=0$.

However, I seem to be stuck now with this equation set:

$\frac{1}{y+z}-\frac{y}{(x+z)^2}-\frac{z}{(y+x)^2}= \lambda$

$\frac{1}{x+z}-\frac{x}{(y+z)^2}-\frac{z}{(y+x)^2}= \lambda$

$\frac{1}{x+y}-\frac{x}{(y+z)^2}-\frac{y}{(z+x)^2}= \lambda$

any help?

Answer 1

$$\frac{x}{y+z}=\frac{x+y+z}{y+z}-1$$

So we get $$LHS=\frac{2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})}{2}-3 \ge \frac{9}{2}-3$$

Using AM-GM inequality, $$LHS \ge \frac{9}{2}-3=\frac{3}{2}$$

Answer 2

On the surface $S:=\{(x,y,z)\,|\, x,y,z >0, \ x+y+z=1\}$ the two functions $f(x,y,z)$ and $$g(x,y,z):={x\over1-x}+{y\over 1-y}+{z\over1-z} $$ coincide. Since the auxiliary function $$t\mapsto \phi(t):={t\over 1-t}=-1+{1\over 1-t}\qquad(t<1)$$ (a hyperbolic arc) is convex Jensen's inequality tells us that for any three real numbers $x$, $y$, $z<1$ we have $${1\over3} g(x,y,z)={1\over3}\bigl(\phi(x)+\phi(y)+\phi(z)\bigr)\geq\phi\left({x+y+z\over3}\right)\ .$$ In the case at hand $x+y+z=1$, so that we obtain $$f(x,y,z)=g(x,y,z)\geq 3\phi\left({1\over3}\right)={3\over2}\ .$$ Note that Lagrange's method only produces the conditionally stationary point $\left({1\over3},{1\over3},{1\over3}\right)$, but you are not told whether $f$ (or $g$) actually takes its global minimum there. Such a minimum need not exist at all, since $S$ is not compact. It could have been that the three points $\bigl(0,{1\over2},{1\over2}\bigr)$, $\ldots$ on the boundary of $S$ lead to a smaller value of $f$ than the point $\left({1\over3},{1\over3},{1\over3}\right)$ in the center of $S$.