Simil convexity inequality

Question

Prove the following inequality if it is true or offer a counterexample if it doesn't hold: $$ \frac{p x + (1-p)y}{pa+(1-p)b} \ge p \frac{x}{a} + (1-p) \frac{y}{b}, $$ with $0\le p \le 1$, $x \ge a \ge 0$, $y \ge b \ge 0$, and $p, a, b, x, y \in \mathbb{R}$.

Answer 1

First, we need $a>0,b>0$ for the right-hand side to be meaningful. Fix $a,b,\text{ and }p$ and define $$f(x,y)=\frac{p x + (1-p)y}{pa+(1-p)b} - p \frac{x}{a} - (1-p) \frac{y}{b}$$ The contention is that $f$ is never negative. We note that $f(x,y)=0$ when $x=a,y=b$. We have $$f_x(x,y)={p\over pa+(1-p)b}-{p\over a}={p(1-p)(a-b)\over{pa^2+(1-p)b}}$$ which is negative when $a<b, 0<p<1$.

So we try $p=\frac12,x=y=b=2,a=1.$ This makes the left-hand side of of the inequality in the question $\frac43$ and the right-hand side $\frac32$, so the statement is false.

Answer 2

Hint

It is equivalent to the convexity of the bivariate function $f(x,y)={x\over y}$ over $\{x\ge y\ge 0\}$. Check the Hessian matrix for that.

Answer 3

After some algebra steps i got this here $${\frac {p \left( a-b \right) \left( ya-xb \right) \left( -1+p \right) }{ \left( pa-bp+b \right) ab}} \geq 0$$