Tricky problem(Old book ex:1.75): (BOMC)

[sourav das]

Given: $a,b,c > 0$ ; $a+b+c=3$ ; $abc=1$

Prove that,$\frac{ab}{(a^2+b)(a+b^2)}+\frac{bc}{(b^2+c)(b+c^2)}+\frac{ca}{(c^2+a)(c+a^2)}\leq\frac{3}{4}$

[zadid xcalibured]

this is the equality case of AMGM so a=b=c=1 and we r done.

[*Mahi*]

Solution:

$\frac {ab} {(a^2+b)(a+b^2)} = \frac {ab} {a^3+a^2b+ab^2+b^3} =\frac 1 {a+b+\frac {a^2} b +\frac {b^2} a}$
Now, by A.M.-G.M.,
$\frac 1 {a+b+\frac {a^2} b +\frac {b^2} a} \leq \frac 1 {4\sqrt {ab}}= \frac {\sqrt c} 4$
Summing, $\frac{ab}{(a^2+b)(a+b^2)}+\frac{bc}{(b^2+c)(b+c^2)}+\frac{ca}{(c^2+a)(c+a^2)} \leq \sum_{cyc} \frac {\sqrt a} 4$
But, by Q.M.-A.M. inequality,
$\sqrt {\frac {a+b+c}{3}} \geq \frac {\sqrt a+\sqrt b+\sqrt c}{3}$
Or,$\sqrt a+\sqrt b+\sqrt c \leq 3$
So, $\frac{ab}{(a^2+b)(a+b^2)}+\frac{bc}{(b^2+c)(b+c^2)}+\frac{ca}{(c^2+a)(c+a^2)} \leq \frac {3} 4$.

[*Mahi*]

I'd love to see the trick though...

[sourav das]

Zadid has given the trick... (Sorry as i have approved the post so lately.) Welcome Zadid...

Trick was in given conditions. For the given condition $a=b=c=1$. Direct equality....

[sourav das]

Solution:

$\frac {ab} {(a^2+b)(a+b^2)} = \frac {ab} {a^3+a^2b+ab^2+b^3} =\frac 1 {a+b+\frac {a^2} b +\frac {b^2} a}$

Factorizing silly mistakes....

[*Mahi*]

The proof still holds in that way... it's about the degree...not what the form is...

[FahimFerdous]

Ha ha ha! Really a nice one!

[*Mahi*]

And whatever, YOU should have seen that the factorization mistake doesn't matter at all, the part of taking the GM is all about the degree of the polynomial.

[bristy1588]

Problem 1.69 of Old book:

Can anyone just show me the steps? How do we apply Rearrangement here?