Prove Inequality for $a,\,b,\,c$: $9abc\ge7(ab+bc+ca)-2$

[anemone]

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

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

 

[MarkFL]

My solution:

Spoiler

Let the objective function be:

\(\displaystyle f(a,b,c)=9abc-7(ab+bc+ca)+2\)

Using my old friend, cyclic symmetry, we find that the extremum occurs for:

\(\displaystyle a=b=c=\frac{1}{3}\)

And we then find:

\(\displaystyle f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\frac{1}{3}-\frac{7}{3}+2=0\)

To ensure this is the minimum, we may look at:

\(\displaystyle f\left(1,0,0\right)=0-0+2=2\)

Thus, we have proved:

\(\displaystyle 9abc\ge7(ab+bc+ca)-2\)

where:

\(\displaystyle a+b+c=1\)

 

[anemone]

My solution:

Spoiler

Let the objective function be:

\(\displaystyle f(a,b,c)=9abc-7(ab+bc+ca)+2\)

Using my old friend, cyclic symmetry, we find that the extremum occurs for:

\(\displaystyle a=b=c=\frac{1}{3}\)

And we then find:

\(\displaystyle f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\frac{1}{3}-\frac{7}{3}+2=0\)

To ensure this is the minimum, we may look at:

\(\displaystyle f\left(1,0,0\right)=0-0+2=2\)

Thus, we have proved:

\(\displaystyle 9abc\ge7(ab+bc+ca)-2\)

where:

\(\displaystyle a+b+c=1\)

Well done MarkFL! And I knew you would tackle it with the help of your old friend! Hehehe...I can read your mind!

Spoiler

I'd buy Mountain Dew for you((Tongueout)) if you try the problem using the Schur's inequality that says, for all non-negative real $x,\,y$ and $z$, and a positive number $t$, we have:

$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge 0$

Hint:

Spoiler

Try $t=1$.

 

[anemone]

I didn't realize until just now that I haven't posted my solution to this old challenge...sorry about that!

Spoiler

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.).