Prooove it!

[sakibtanvir]

a,b,c are positive real numbers.\[abc=1\]Prove that,\[\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq \frac{3}{2}\]

[nafistiham]

WLOG let us assume $c\leq b\leq a$ let $x= \frac {1}{a},y=\frac {1}{b},z=\frac {1}{c}$ now,
now, we can find the left hand side,
\[\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{x}+\frac{1}{z}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}}\]
\[=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\]
since $x\leq y \leq z$ we can deduce that, $x+y\leq x+z \leq y+z$ and also that, $\frac{x}{y+z}\leq \frac{y}{x+z}\leq \frac{z}{x+y}$
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xy}{y+z}+ \frac{yz}{x+z}+ \frac{zx}{x+y}\]
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xz}{y+z}+ \frac{yx}{x+z}+ \frac{zy}{x+y}\]
if we add both sides we'll get
\[2 \left ( \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right) \geq x+y+z \geq 3\sqrt[3]{xyz} = 3\]
therefore,
\[\frac{1}{a^{3}(b+c)}+\frac{1} {b^{3}(c+a)}+\frac{1}{c^{3}(b+c)} \geq \frac{3}{2} \]

[Tahmid Hasan]

এইটা একটা আই এম ও প্রব্লেম আর আগেও এটি পোস্ট হয়েছে।

[Nadim Ul Abrar]

$ab+bc+ca\geq 3\sqrt{abc}=3$ [AM-GM]

$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}=\frac{(bc)^2}{a(b+c)}+\frac{(ca)^2}{b^3(c+a)}+\frac{(ab)^2}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2} \geq \frac{3}{2} $[ C.S Engel ]

[sakibtanvir]

WLOG let us assume $c\leq b\leq a$ let $x= \frac {1}{a},y=\frac {1}{b},z=\frac {1}{c}$ now,
now, we can find the left hand side,
\[\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{x}+\frac{1}{z}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}}\]
\[=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\]
since $x\leq y \leq z$ we can deduce that, $x+y\leq x+z \leq y+z$ and also that, $\frac{x}{y+z}\leq \frac{y}{x+z}\leq \frac{z}{x+y}$
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xy}{y+z}+ \frac{yz}{x+z}+ \frac{zx}{x+y}\]
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xz}{y+z}+ \frac{yx}{x+z}+ \frac{zy}{x+y}\]
if we add both sides we'll get
\[2 \left ( \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right) \geq x+y+z \geq 3\sqrt[3]{xyz} = 3\]
therefore,
\[\frac{1}{a^{3}(b+c)}+\frac{1} {b^{3}(c+a)}+\frac{1}{c^{3}(b+c)} \geq \frac{3}{2} \]

I didn't know it was posted before...by the way..thanks to help.bro. Can i know how much time did u spend to prove it?

[nafistiham]

well, actually i did it during the inequality camp going on the group some days ago.as, the task was to learn most part of book "Inequalities", i used the examples to do the exercises.
and this is one of the solved examples of that book.

[Phlembac Adib Hasan]

Another proof:

\[\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{1}{a^2(ab+ac)}\]
Now let's apply Holder's inequality[sorry for wrong spelling] to this expression:
\[(\sum_{cyc}\frac{1}{a^2(ab+ac)})(\sum_{cyc}\frac{1}{a})(\sum_{cyc}ab+ac)\geq (\sum_{cyc}\frac{1}{a})^3\]
\[\Rightarrow (\sum_{cyc}\frac{1}{a^2(ab+ac)})\; \; \; 2.(\sum_{cyc}\frac{1}{a})^2\geq (\sum_{cyc}\frac{1}{a})^3\]
\[\Rightarrow 2(\sum_{cyc}\frac{1}{a^2(ab+ac)})\geq (\sum_{cyc}\frac{1}{a})\geq \frac{3}{\sqrt[3]{abc}}\; \; \; [AM-GM]\]
\[\Rightarrow (\sum_{cyc}\frac{1}{a^2(ab+ac)})\geq \frac{3}{2}\]