For reals $a,\,b,\,c\ge 0$, prove the inequality

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

It's hard not to relate the LHS of the inequality with the AM-GM inequality formula, but if we persist, we would apply the AM-GM inequality to the LHS, and we get:

$\begin{align*}a(a-c)^2+b(b-c)^2&\ge 2\sqrt{a(a-c)^2b(b-c)^2}\\&\ge 2(a-c)(b-c)\sqrt{ab}\end{align*}$

If we're heading in the right direction, we would definitely get the following inequality that holds for all $a,\,b,\,c\ge 0$.

$2(a-c)(b-c)\sqrt{ab}\ge (a-c)(b-c)(a+b-c)$

The above is true if $a=b=c=0$ or $a=b=c\ne 0$.

If $a\ne b\ne c$, we need to prove $2\sqrt{ab}\ge (a+b-c)$ also holds for all positive real $a,\,b,\,c$.

$2(a-c)(b-c)\sqrt{ab}\ge (a-c)(b-c)(a+b-c)$

$2\sqrt{ab}\ge (a+b-c)$

Squaring both sides of the inequality:

$4ab\ge (a+b)^2-2c(a+b)+c^2$

$4ab\ge a^2+2ab+b^2-2ac-2bc+c^2$

$2ab+2bc+2ac\ge a^2+b^2+c^2$

Replace $2ab+2bc+2ac$ by $2ab+2bc+2ac=(a+b+c)^2-(a^2+b^2+c^2)$, the above inequality becomes

$(a+b+c)^2-(a^2+b^2+c^2)\ge a^2+b^2+c^2$

$(a+b+c)^2\ge 2(a^2+b^2+c^2)$---(*)

But from the Cauchy–Schwarz inequality, we have:

$\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}\ge ab+bc+ac$

$a^2+b^2+c^2\ge ab+bc+ac$

$2(a^2+b^2+c^2)\ge 2(ab+bc+ac)$

$2(a^2+b^2+c^2)+a^2+b^2+c^2\ge a^2+b^2+c^2+2(ab+bc+ac)$

$3(a^2+b^2+c^2)\ge (a+b+c)^2$---(**)

The result from (**) doesn't suggest (*) must be correct.

Thus we're back to square one.

I encourage you to try the problem and I'll lead you to the proof on my next blog post.

## No comments:

## Post a Comment