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Tuesday, June 7, 2016

Prove the following inequality holds: (log3ab+log3ac)+(log3bc+log3ba)+(log3ca+log3cb)36.

Let the reals a,b,c(1,) with a+b+c=9.

Prove the following inequality holds:

(log3ab+log3ac)+(log3bc+log3ba)+(log3ca+log3cb)36.

My solution:

By applying the Cauchy-Schwarz inequality to the LHS of the intended inequality gives
(log3ab+log3ac)+(log3bc+log3ba)+(log3ca+log3cb)

1+1+1log3ab+log3ac+log3bc+log3ba+log3ca+log3cb

=3log3aa+log3ab+log3ac+log3ba+log3bb+log3bc+log3ca+log3cb+log3cclog3aalog3bblog3cc

\displaystyle =\sqrt{3}\sqrt{\log_3a^{a+b+c} +\log_3b^{a+b+c} +\log_3c^{a+b+c} -(\log_3a^a+\log_3b^b+\log_3c^c})

\displaystyle =\sqrt{3}\sqrt{(a+b+c)\log_3abc -(\log_3a^a+\log_3b^b+\log_3c^c})

\displaystyle =\sqrt{3}\sqrt{9\log_33^3 -(\log_3a^a+\log_3b^b+\log_3c^c}) since \displaystyle a+b+c=9\ge 3\sqrt[3]{abc}\implies abc \le 3^3

\displaystyle =\sqrt{3}\sqrt{27 -(a\log_3a+b\log_3b+c\log_3c})

Now, if we're to study the nature of the function for \displaystyle f(a)=a\log_3 a, we know it's a concave up and an increasing function, we could use the Jensen's inequality to figure out the minimum of \displaystyle a\log_3a+b\log_3b+c\log_3c:

\displaystyle \frac{a\log_3a+b\log_3b+c\log_3c}{3}\ge \frac{a+b+c}{3}\log_3\left(\frac{a+b+c}{3}\right)=\frac{9}{3}\log_3\left(\frac{9}{3}\right)=3

\displaystyle \therefore a\log_3a+b\log_3b+c\log_3c=3(3)=9

Now we get the maximum of the LHS of the intended inequality as:

\displaystyle \begin{align*}\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}&\le \sqrt{3}\sqrt{27 -(a\log_3a+b\log_3b+c\log_3c})\\& \le \sqrt{3}\sqrt{27 -9}\\&=\sqrt{3}\sqrt{18}\\&=\sqrt{9}\sqrt{6}\\&=3\sqrt{6}\end{align*}

with equality when \displaystyle a=b=c=3.

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