Let $a,\,b$ and $c$ be positive real numbers satisfying $a+b+c=1$. Prove that $9abc\ge7(ab+bc+ca)-2$.

Let $a,\,b$ and $c$ be positive real numbers satisfying $a+b+c=1$.

Prove that $9abc\ge7(ab+bc+ca)-2$.

My solution:

Schur's inequality says, for all positive real $a,\,b$ and $c$, we have:

$9abc\ge 4(ab+bc+ca)(a+b+c)-(a+b+c)^3$(*)

In our case, we're given $a+b+c=1$, so substituting that into (*) yields

$9abc\ge 4(ab+bc+ca)-1$(**)

If we can prove $4(ab+bc+ca)-1\ge 7(ab+bc+ca)-2$, then we're done.

From the Cauchy-Schwarz inequality, we have:

$a^2+b^2+c^2\ge ab+bc+ca\implies (a+b+c)^2\ge 3(ab+bc+ca)$

Therefore, with $a+b+c=1$, the above inequality becomes

$1\ge 3(ab+bc+ca)$

Add the quantity $4(ab+bc+ca)-2$ to both sides we obtain:

$4(ab+bc+ca)-2+1\ge 4(ab+bc+ca)-2+3(ab+bc+ca)$

$4(ab+bc+ca)-1\ge 7(ab+bc+ca)-2$ (Q.E.D.)