Question –
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$$
My try
$$
\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
$$
which implies that
$$
\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}}
$$
It suffices to prove that
$$
(a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right.
$$
beacuse we can easily get that $\sum a \ge 3$
but i am not able to prove it..
note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet…
any hints ???
thankyou
Best Answer
It is enough to show that $\sqrt a + \sqrt b + \sqrt c = 3 \implies (ab)^{2/3} + (bc)^{2/3} + (ca)^{2/3} \leqslant 3$. For ease, let us replace $a, b, c$ with $x^6, y^6, z^6$, so we need to show for positives, $x^3+y^3+z^3=3 \implies (xy)^4+(yz)^4+(zx)^4\leqslant 3$. This one is in fact a known old chestnut.
We note by AM-GM, $xy \leqslant \frac13(x^3+y^3+1) = \frac13(4-z^3)$, hence $(xy)^4 \leqslant \frac13(4x^3y^3-x^3y^3z^3)$. Cyclically summing three such inequalities, we get $$\sum (xy)^4 \leqslant \frac43\sum (xy)^3-(xyz)^3$$ Now with $X=x^3, Y = y^3, Z = z^3$, it is enough to show with $X+Y+Z=3$ $$4(XY+YZ+ZX) -3XYZ\leqslant 9$$ which is the well known Schur's inequality: $$4(X+Y+Z)(XY+YZ+ZX) \leqslant (X+Y+Z)^3+9XYZ$$