If $a^2+b^2+c^2=1$ prove $\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$

Question

Let $a,b,c\ge 0: a^2+b^2+c^2=1.$ Prove that $$\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$$

Here is just my thought:

After clear denominator, it suffices to prove $$\sum_{cyc}(a^4-a^2)(ac-1)(ab-1)\le (ab+bc+ca)\prod_{cyc}(bc-1)$$which $uvw$ might be useful.

I tried to use pqr but it is complicated.

Also, the equality point $$\left(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\right);\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0\right);(1,0,0)$$ is a trouble.

I hope to see proofs which can be verified by hand and attained it in during competition.

Thank you for helping.

Answer 1

Proof.

By using AM-GM and Cauchy-Schwarz \begin{align*} \frac{a^4-a^2}{bc-1}&=\frac{a^2(b^2+c^2)}{a^2+b^2+c^2-bc}\\&\le \frac{a^2(b^2+c^2)}{2a\sqrt{b^2+c^2-bc}}\\&=\frac{a(b+c)(b^2+c^2)}{2\sqrt{(b+c)(b^3+c^3)}}\\&\le\frac{a(b+c)(b^2+c^2)}{2(b^2+c^2)}\\&=\frac{ab+ac}{2}. \end{align*} Sum up similar inequalities, the desired result follows.

Answer 2

Because $$ab+ac+bc-\sum_{cyc}\frac{a^4-a^2}{bc-1}=\sum_{cyc}\left(\frac{ab+ac}{2}-\frac{a^2(b^2+c^2)}{a^2+b^2+c^2-bc}\right)=$$ $$=\sum_{cyc}\tfrac{a(a^2(b+c)+b^3+c^3)-2a(b^2+c^2))}{2(a^2+b^2+c^2-bc)}=\sum_{cyc}\tfrac{a(b(a-b)^2+c(a-c)^2)}{2(a^2+b^2+c^2-bc)}\geq0.$$

Answer 3

Note that $ab+bc+ca \le a^2+b^2+c^2 =1$. Also, $a,b,c \le 1$ So, \begin{align*} \sum_{cyc} \dfrac{a^4-a^2}{bc-1} = & \sum_{cyc} \dfrac{a^2(1-a^2)}{1-bc} \\ \le & \sum_{cyc} \dfrac{a^2(b^2+c^2)}{ab+ac} \\ = & \sum_{cyc}a \left(b+c -\dfrac{2bc}{b+c}\right) \\ = & 2 \left(\sum_{cyc} ab\right) - \left( \sum_{cyc} \dfrac{2bc}{b+c} \right) \\ = & ab+bc+ca + \sum_{cyc} ab \left( 1-\dfrac2{a+b} \right) \\ \le & ab+bc+ca \end{align*}

Hope, this helps you.