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

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The inequality to prove is that for positive real numbers \(a\), \(b\), and \(c\) with \(a+b+c=1\), the expression \(9abc\) is greater than or equal to \(7(ab+bc+ca)-2\). Participants in the discussion share their solutions and insights, with one individual acknowledging the contributions of another. There is a light-hearted exchange about the problem-solving process. The conversation highlights the collaborative nature of tackling mathematical challenges. Overall, the focus remains on proving the stated inequality.
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Let $a,\,b$ and $c$ be positive real numbers satisfying $a+b+c=1$.

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

Let the objective function be:

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

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

$$a=b=c=\frac{1}{3}$$

And we then find:

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

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

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

Thus, we have proved:

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

where:

$$a+b+c=1$$
 
MarkFL said:
My solution:

Let the objective function be:

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

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

$$a=b=c=\frac{1}{3}$$

And we then find:

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

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

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

Thus, we have proved:

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

where:

$$a+b+c=1$$

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

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:

Try $t=1$.
 
I didn't realize until just now that I haven't posted my solution to this old challenge...sorry about that!:o

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.).
 
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