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

Question

Let $a,b$ and $c$ be positive reals such that $a+b+c=3$. Prove that $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq a^2+b^2+c^2 $$

Attempt : Homogenizing the equation by letting $a=\frac{3x}{x+y+z}, b=\frac{3y}{x+y+z}$ and $c=\frac{3z}{x+y+z}$, the inequality becomes $$\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\geq \frac{3\left(x^2+y^2+z^2\right)}{x+y+z}$$ where $x,y,z >0$. I am stuck here. Just a hint would suffice.(No calculus)

Answer 1

Let $c=\min\{a,b,c\}$.

Hence, we need to prove that $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq\frac{3(a^2+b^2+c^2)}{a+b+c}$$ or $$\frac{a^3}{b^2}+\frac{b^3}{a^2}-a-b+\frac{b^3}{c^2}+\frac{c^3}{a^2}-\frac{b^3}{a^2}-c\geq\frac{3(a^2+b^2+c^2)}{a+b+c}-a-b-c$$ or $$\frac{(a-b)^2(a+b)(a^2+ab+b^2)}{a^2b^2}+\frac{(c-a)(c-b)(a+c)(b^2+bc+c^2)}{a^2c^2}\geq$$ $$\geq\frac{2(a-b)^2}{a+b+c}+\frac{2(c-a)(c-b)}{a+b+c},$$ which is obvious.

Done!

Answer 2

Here is my solution: Assume $x\geq y\geq z>0$ $$\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\geq \frac{3\left(x^2+y^2+z^2\right)}{x+y+z}$$ or $$\left(\frac{x^4}{y^2}+\frac{y^4}{z^2}+\frac{z^4}{x^2}\right)+\left(\frac{xy^3}{z^2}+\frac{x^3}{y}\right)+\left(\frac{yz^3}{x^2}+\frac{y^3}{z}\right)+\left(\frac{zx^3}{y^2}+\frac{z^3}{x}\right)\geq 3\left(x^2+y^2+z^2\right)$$ Now $$\frac{x^4}{y^2}+\frac{y^4}{z^2}+\frac{z^4}{x^2}\geq x^2+y^2+z^2\,\,\,\text{(Cauchy–Schwarz)}$$ and $$\left(\frac{xy^3}{z^2}+\frac{x^3}{y}\right)+\left(\frac{yz^3}{x^2}+\frac{y^3}{z}\right)+\left(\frac{zx^3}{y^2}+\frac{z^3}{x}\right)\geq \frac{2x^2y}{z}+\frac{2y^2z}{x}+\frac{2z^2x}{y}\,\text{(AM-GM)}$$ So, it is enough to prove that $$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\geq x^2+y^2+z^2$$ which after simplifying becomes $$y(x-z)(y-z)(x^2+zx-zy)+z^3(x-y)^2\geq 0$$ which follows from the assumption.