$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

Question

Question

Let.a, $b, c$ be positive real numbers with sum 3 . Prove that $$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $$

my doubt -

by using cauchy reverse technique i have estimate the given expression with difference of two another expression and i just want to prove that

$b \sqrt[3]{a^{2}}+c \sqrt[3]{b^{2}}+a \sqrt[3]{c^{2}} \leq 3$

now they write

According to AM-GM, we obtain $$ 3 \sum_{c y c} a \geq \sum_{c y c} a+2 \sum_{c y c} a b=\sum_{c y c}(a+a c+a c) \geq 3 \sum_{c y c} a \sqrt[3]{c^{2}} $$

but how they proved that $ 3 \sum_{c y c} a \geq \sum_{c y c} a+2 \sum_{c y c} a b$

this means that $a+b+c > ab+bc+ca$ how ???

i know this is little doubt but i want to clear it ....

thankyou

Answer 1

Hint: AM-GM

$3 \sum a = (\sum a )^2 = (a^2+b^2+c^2) + 2\sum ab = \left(\frac{1}{2} \sum ( a^2 + b^2 ) \right) + 2 \sum ab $
$ \geq \sum ab + 2 \sum ab = 3 \sum ab $

Answer 2

Another way.

We need to prove that $$a^3c^2+b^3a^2+c^3b^2\leq3$$ for non-negatives $a$, $b$ and $c$ such that $a^2+b^2+c^2=3$.

Indeed, let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.

Thus, by Rearrangement, AM-GM and AM-GM we obtain: $$a^3c^2+b^3a^2+c^3b^2=a\cdot a^2c^2+b\cdot b^2a^2+c\cdot c^2b^2\leq x\cdot x^2y^2+y\cdot x^2z^2+z\cdot y^2z^2=$$ $$=y(x^3y+x^2z^2+z^3y)=y\left(x^2\left(xy+\frac{z^2}{2}\right)+z^2\left(\frac{x^2}{2}+yz\right)\right)\leq$$ $$\leq y\left(x^2\left(\frac{x^2+y^2}{2}+\frac{z^2}{2}\right)+z^2\left(\frac{x^2}{2}+\frac{y^2+z^2}{2}\right)\right)=\frac{3}{2}y(3-y^2)\leq3.$$