RMO 2016

Relevant wiki: Titu's Lemma

The $LHS$ is equivalent to $\displaystyle\sum_{cyc}\frac{a^2}{a+ab}$ , by Titu's Lemma $1=\displaystyle\sum_{cyc}\frac{a^2}{a+ab}\geq\frac{(a+b+c)^2}{a+b+c+ab+bc+ca}$ $\Leftrightarrow a+b+c+ab+bc+ca\geq (a+b+c)^2$ Using $ab+bc+ca\stackrel{C-S}\leq\frac{(a+b+c)^2}{3}$ , we have $a+b+c+\frac{(a+b+c)^2}{3}\geq (a+b+c)^2$ $\therefore \frac{3}{2}\geq a+b+c\stackrel{AM-GM}\geq 3\sqrt[3]{abc}$ $\Leftrightarrow abc\leq\frac{1}{8}=0.125$ The equality holds when $a=b=c=\frac{1}{2}$

Given :

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

Taking LCM and cross multiplying we get:

$a\cdot(1+c)\cdot(1+a) + b\cdot(1+b)\cdot(1+a) + c\cdot(1+b)\cdot(1+c) = (1+a)\cdot(1+b)\cdot(1+c)$

After simplifying and cancelling the like terms on both sides we get:

$1+ abc = a^{2}+ b^{2} + c^{2}+ a^{2}c + b^{2}a + c^{2}b$

Using AM GM inequality separately on the terms: $( a² + b² + c² )$ and $(a²c + b²a + c²b)$ we get:

$1 + abc \ge 3(abc)^{2/3}+ 3abc$

Equality occurs when $a = b = c = 1/2$

Now, cubing both sides of inequality we get

$(1 - 2abc )³ \ge 27a²b²c²$

Let $abc = t$

$1 - 8t³ - 6t + 12t² \ge 27t²$

$8t³ + 15t² + 6t - 1 \le 0$

The above expression can also be written as:

$8t³ + 16t² + 8t - t² - 2t - 1 \le 0$

$8t ( t² + 2t + 1) - 1( t² + 2t + 1) \le 0$

$(8t - 1)(t+1)² \le 0$

So $8t - 1 \le 0$ as $( t + 1)² >0$

$8t \le 1$

$8abc \le 1$

$abc \le 1/8 = 0.125$

Hence the maximum value of $abc$ is $0.125$

Try my problem ' A cubic inequality'

Using the Cauchy-Schwartz inequality, we have $\bigg((1+b)+(1+c)+(1+a)\bigg)\bigg( \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}\bigg)\geq \big(\sqrt{a}+\sqrt{b}+\sqrt{c}\big)^2=a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$ Using the AM-GM inequality on the RHS, we have from the above $3+(a+b+c)\geq a+b+c+ 6(abc)^{1/3},$ which implies $abc\leq \frac{1}{8}.$ The equality is achieved when $a=b=c=\frac{1}{2}.$

From AM,GM we can easily Conclude that maximum value Of abc, Will be when a,b,c are are equal. Now take a= b=c and Solve it.