# Prove that $a^2x + b^2y \geq 2abc$, where $c \leq x, c \leq y$

Prove that $a^2x + b^2y \geq 2abc$, where $c \leq x, c \leq y$.

I've proved that $a^2 + b^2 \geq 2ab$, but I don't know how to deal with the coefficient terms. The question seems logical, given that $c$ is less than both $x$ and $y$, but I want to prove this more rigorously.

• are the numbers $a,b,c$ are assumed to be positive? – Dr. Sonnhard Graubner Nov 8 '17 at 20:15
• Sorry I should add more details. The question actually relates to a probability proof, which I've reduced to this question. $a$, $b$ are any real numbers, while x, y, and c range from 0 to 1 (they are probabilities). – hedebyhedge Nov 8 '17 at 20:18

Just substitute what you already have, if you replace $a$ and $b$ in your formula with $a\sqrt{x}$ and $b\sqrt{y}$ and use than $c^2\leq{xy}$ implying $c \leq{\sqrt{xy}}$

if so then by $AM-GM$ we have $$a^2x+b^2y\geq 2ab\sqrt{xy}$$ and from $$x\geq c$$ and $$y\geq c$$ we get $$xy\geq bc^2$$ thus we have $$2ab \sqrt{xy}\geq 2ab\sqrt{c^2}=2abc$$

From $a^2 + b^2 \geq 2ab$ you can get $a^2 -2ab + b^2 \geq 0$ which can be factored into $(a-b)^2 \geq 0$ square root both sides gives you $a-b \geq 0$ then $a \geq b$, so as long as thats true the previos is also true.

If $x \ge c \ge 0$ and $y \ge c \ge 0$. Then if $k \ge m$ then $k*x \ge k*c \ge m*c$.

So if $a^2 + b^2 \ge 2ab$ (by AM-GM or completing the square).

then $a^2x + b^2y \ge a^2c + b^2c \ge 2abc$

Let $x=c+u$ and $y=c+v$.

Thus, $u\geq0$, $v\geq0$ and $$a^2x+b^2y-2abc=a^2(c+u)+b^2(c+v)-2abc=c(a-b)^2+a^2u+b^2v\geq0.$$