Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\leq 1$.

Question

Let $a, b, c>0$ s.t. $abc (a+b+c)=3$.

Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+ \frac{1}{c^2+a^2+1}\leq 1$.

I have no idea how to start.

Answer 1

By C-S we obtain: $$\sum_{cyc}\frac{1}{a^2+b^2+1}=\sum_{cyc}\frac{2+c^2}{(a^2+b^2+1)(1+1+c^2)}\leq\sum_{cyc}\frac{2+c^2}{(a+b+c)^2}.$$ Can you end it now?

Answer 2

Firstly, we have $abc(a+b+c)=3$, which means that the known inequality $abc \le \frac{(a+b+c)^3}{27}$ gives $\frac{3}{a+b+c}\le\frac{(a+b+c)^3}{27}$, giving further $a+b+c \ge 3$.

And, therefore, $abc = \frac{3}{a+b+c} \le 1$, which we shall call inequality (1)

We know $ \sum_{cyc}{a^2} \ge \frac{(a+b+c)^2}{3} \ge 3 $, which we shall call inequality (2)

Making it such that you have the same denominator everywhere, the inequality will become equivalent to $$2 \sum {a^2} +2 \le \sum {a^2b^2(a^2+b^2)} + 2a^2b^2c^2 $$ $$\text {or, equivalently,}$$ $$2 \sum_{cyc} {a^2} +2 + a^2b^2c^2 \le (\sum_{cyc}{a^2})(\sum_{cyc} {(ab)^2}) $$ which is true since,$$\text{ by} \sum_{cyc}{x^2} \ge \sum_{cyc}{xy} \text{ applied for the last cyclical sum, where } (x,y,z)=(ab,bc,ac) \text{ we have}$$ $$\sum_{cyc}{a^2}\sum_{cyc}{(ab)^2} \ge (\sum_{cyc}{a^2})[abc(a+b+c)] = 3\sum_{cyc}{a^2} = 2\sum_{cyc}{a^2} + \sum_{cyc}{a^2} \ge $$ $$ 2\sum_{cyc}{a^2}+3 \ge 2\sum_{cyc}{a^2}+ 2 + a^2b^2c^2 $$ The two last steps were made using inequality (1), respectively (2), which we know to be true. Q.E.D.