prove $\frac{x}{(ay + bz)} + \frac{y}{(az + bx)} + \frac{z}{(ax + by)} ≥ \frac{3}{(a+b)}$

Question

Let a, b, x, y, z be positive real numbers.
How is it possible to prove the following inequality:

$$\frac{x}{(ay + bz)} + \frac{y}{(az + bx)} + \frac{z}{(ax + by)} ≥ \frac{3}{(a+b)}$$

Answer 1

By Cauchy Schwarz we get $$\frac{x}{ay+bz}+\frac{y}{az+by}+\frac{z}{ax+by}=\frac{x^2}{axy+bxz}+\frac{y^2}{ayz+bxy}+\frac{z^2}{axz+byz}\geq \frac{(x+y+z)^2}{axy+bxz+ayz+bxy+axz+byz}$$ Now we have to show that $$\frac{(x+y+z)^2}{axy+bxz+ayz+bxy+axz+byz}\geq \frac{3}{a+b}$$ This is true, since it is equivalent to $$(x^2+y^2+z^2-xy-yz-zx)^2(a+b)\geq 0$$ which is true.

Answer 2

By C-S $$\sum_{cyc}\frac{x}{ay+bz}=\sum_{cyc}\frac{x^2}{axy+bxz}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(axy+bxz)}=\frac{(x+y+z)^2}{\sum\limits_{cyc}(axy+bxy)}=$$ $$=\frac{\sum\limits_{cyc}(x^2+2xy)}{(a+b)\sum\limits_{cyc}xy}\geq\frac{\sum\limits_{cyc}(xy+2xy)}{(a+b)\sum\limits_{cyc}xy}=\frac{3}{a+b}.$$

Answer 3

WLOG assume $a+b = 1$. Put $m = \dfrac{y}{x}, n = \dfrac{z}{x}, p = \dfrac{z}{y}, q = \dfrac{x}{y}, r = \dfrac{x}{z},s = \dfrac{y}{z}\implies mq = nr = ps = 1\implies LHS = \dfrac{1}{am+bn}+\dfrac{1}{ap+bq}+\dfrac{1}{ar+bs}=f(a,b)$. Observe that $f$ is convex in $a$ and $b$ , thus the extreme values of $f$ must be attained at the end points which are $(a,b) = (0^{+}, 1^{-}), (1^{-}, 0^{+})$, and $f(0^{+},1^{-}) = \dfrac{1}{n}+\dfrac{1}{q}+\dfrac{1}{s}\ge 3\sqrt[3]{\dfrac{1}{nqs}} = 3$ by AM-GM inequality. Similarly, $f(1^{-},0^{+}) \ge 3$. Proving the assertion.