# Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$

Prove that for every $$a,b,c \in \mathbb{R}^{+}$$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$ Unfortunately i can just prove that : $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$ like this : $$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$ and by the same method we have : $$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$ Adding them together will give us the desired inequality.

and please don't use any $$\sum_{cyc}$$ because I get confused with it.

Suppose $$a \geqslant b \geqslant c,$$ we have $$\frac{a^2}{a^2+bc} < 1,$$ and $$\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} \leqslant \frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+b^2} = 1.$$ Therefore $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} < 2 .$$

• Is it always helpful to suppose that $a\geq b\geq c$ If there is a symmetry in the inequality?
– PNT
Feb 18, 2021 at 9:46
• One small comment: the first inequality is strict, therefore the last inequality is also strict. Feb 18, 2021 at 13:24

We can rewrite the LHS :

$$LHS = \frac1{1+ \frac{bc}{a^2}} + \frac1{1+\frac{ac}{b^2}} + \frac1{1+\frac{ab}{c^2}}$$

Let $$u = \frac{b}{a}$$, $$v = \frac{a}{c}$$, $$w = \frac{c}{b}$$. Then :

$$LHS = \frac{v}{u+v} + \frac{u}{w+u} + \frac{w}{w+v}$$

We suppose that $$u \geq v$$ and $$u \geq w$$ (other cases are similar).

If $$w > v$$ : $$LHS < \frac{v}{u+v} + \frac{u}{v+u} + \frac{w}{w} = 2$$

If $$w \leq v$$ :

$$LHS < \frac{v}{v+v} + \frac{u}{u} + \frac{w}{w+w} = 2$$

Taking $$u = n^2$$, $$w = n$$, $$v = 1$$ :

$$LHS = \frac1{1+n}+\frac{n^2}{n+n^2}+\frac{n}{n+1} \to 2 \quad [n \to \infty]$$

Thus $$2$$ can't be improved.

let $$p=\frac{a^2}{a^2+bc},q=\frac{b^2}{b^2+ac},r=\frac{c^2}{c^2+ab}$$ then easy to see $$p,q,r\in (0,1)$$ Now notice that $$\frac{(1-p)(1-q)(1-r)}{pqr}=1$$ so $$p+q+r=1+pq+qr+rp-2pqr$$ It remains to prove $$pq+qr+rp-2pqr\le 1$$ $$\iff \underbrace{p(1-q)(r-1)}_{\le 0}+\underbrace{q(1-r)(p-1)}_{\le 0}+\underbrace{(1-p)(q-1)}_{\le 0}\le 0$$ which is obvious as each term is $$\le 0$$ as shown

• Neat Schur's inequality. And change of variables. Feb 18, 2021 at 3:29
• @SketchySketch thanks! but I havent used schur anywhere ,I am missing something? Feb 18, 2021 at 3:59
• sorry I'm missing something :) Feb 18, 2021 at 5:58
• +1 Nice change of variables. For the last step, if finding that split is hard (it wasn't obvious to me), we could also prove it by observing that the expression is linear in $p$ when $q,r$ is fixed, so we just need to check the endpoints. Feb 18, 2021 at 21:38
• @CalvinLin Thanks!Initially(if you see my first post in edit history) I had done that (ie considering it as a lenear function and etc) ,however I spent around 15 mins to find that split and edited my post to incorporate that Feb 19, 2021 at 1:14

From $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2 \iff\\ \frac{a^2+bc-bc}{a^2+bc}+\frac{b^2+ac-ac}{b^2+ac}+\frac{c^2+ab-ab}{c^2+ab}<2 \iff\\ 3-\frac{bc}{a^2+bc}-\frac{ac}{b^2+ac}-\frac{ab}{c^2+ab}<2 \iff\\ \frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}>1$$ and applying Titu's Lemma $$\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}=\\ \frac{b^2c^2}{a^2bc+b^2c^2}+\frac{a^2c^2}{b^2ac+a^2c^2}+\frac{a^2b^2}{c^2ab+a^2b^2}\geq\\ \frac{(bc+ac+ab)^2}{b^2c^2+a^2c^2+a^2b^2+a^2bc+b^2ac+c^2ab}=\\ \frac{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{2a^2bc+2b^2ac+2c^2ab}}{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{a^2bc+b^2ac+c^2ab}}>1$$

Here's the inevitable 'expand and see if everything works out' solution. $$\frac{a^2}{a^2 + bc} + \frac{b^2}{b^2 + ac}+\frac{c^2}{c^2 + ab} < 2 \\ \iff a^2 (b^2 + ac)(c^2 + ab) + b^2 (a^2 + bc)(c^2 + ab) + c^2(a^2 + bc)(b^2 + ac) < 2 (a^2 + bc)(b^2 + ac)(c^2 + ab) \\ \iff 3a^2b^2c^2 + 2(a^3b^3 + b^3c^3 + c^3a^3) + a^4 bc + b^4 ac + c^4ab < 2(2a^2b^2c^2 + a^3 b^3 + a^3c^3 + b^3 c^3 + a^4 bc + b^4ac + c^4 ab)$$ It's important when doing these sort of calculations to make sure you didn't miss a term. Here the number of terms on the left (before summing them up) is $$12 = 3 \times 2^2$$ and the number of terms on the right is $$8 = 2^3$$ as expected.

This inequality simplifies to $$a^2b^2c^2 +a^4 bc + b^4ac + c^4 ab > 0$$ Which is evidently true.