Prove $1< \frac a{\sqrt{a^2+b^2}} + \frac b{\sqrt{b^2+c^2}} + \frac c{\sqrt{c^2+a^2}} \le \frac{3\sqrt2}2$

Question

Suppose that a, b, c are positive real numbers, prove that : $$1 < \frac a{\sqrt{a^2+b^2}} + \frac b{\sqrt{b^2+c^2}} + \frac c{\sqrt{c^2+a^2}} \le \frac{3\sqrt2}2.$$

This is a question from 100 inequalities by Vasc and Arqady. Link

For the Right Hand Side inequality,

I have tried applying AM-GM inequality and Cauchy–Schwarz inequality to solve the inequality but the issue is I am unable to arrange the terms for the $\le$ sign. The terms I want on the left side seem to end up on the right and vice versa. Even if I try to solve using the flipped sign of inequality, I am getting stuck midway unable to simplify the terms.

Terms used in AM-GM inequalty :

$A : \frac a{\sqrt{a^2+b^2}}$

$B : \frac b{\sqrt{b^2+c^2}}$

$C : \frac c{\sqrt{c^2+a^2}}$

and then continuing with $\frac {A+B+C}3 \ge \sqrt[3] {ABC}$ .

For the Cauchy–Schwarz inequality, I am using :

$a_1 : \frac a{\sqrt{a^2+b^2}}, \ \ a_2 : \frac b{\sqrt{b^2+c^2}}, \ \ a_3 : \frac c{\sqrt{c^2+a^2}}$ and

$b_1 : \frac{\sqrt{a^2+b^2}}a, \ \ b_2 : \frac{\sqrt{b^2+c^2}}b, \ \ b_3 : \frac{\sqrt{c^2+a^2}}c$

and then continuing with $(a_1^2+a_2^2+a_3^2)*(b_1^2+b_2^2+b_3^2) \ge (a_1b_1 + a_2b_2 + a_3b_3)^2$

For the Left Hand Side Inequality,

I have the intuition that the inequality holds true but I cannot seem to find a method to prove it.

Any ideas, hints and approaches are welcome. Thank you very much.

Answer 1

For the LHS: $$ \begin{align} \frac a{\sqrt{a^2+b^2}} + \frac b{\sqrt{b^2+c^2}} + \frac c{\sqrt{c^2+a^2}} &> \frac a{\sqrt{a^2+b^2+c^2}} + \frac b{\sqrt{a^2+b^2+c^2}} + \frac c{\sqrt{a^2+b^2+c^2}}\\ &= \frac {a+b+c}{\sqrt{a^2+b^2+c^2}}\\ &> 1 \end{align} $$ where the last inequality follows from $$ a+b+c = \sqrt{(a+b+c)^2} > \sqrt{a^2+b^2+c^2}. $$

Answer 2

For the Right Hand Side Inequality

Let $x=a^2,y=b^2,z=c^2$. Then, we just need to prove: $$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leq \frac{3}{\sqrt{2}}$$ By C-S, we have: $$\sum_{cyc}{\sqrt{\frac{x}{x+y}}} \leq \sqrt{\sum_{cyc}(x+z)\sum_{cyc}\frac{x}{(x+y)(x+z)}}=\sqrt{\frac{4(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(x+z)}} $$ Hence, we just need to prove: $$\frac{4(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(x+z)} \leq \frac{9}{2}$$ $$\Leftrightarrow 9(x+y)(y+z)(x+z) \ge8(x+y+z)(xy+yz+xz)$$ $$\Leftrightarrow x^2y+y^2z+z^2x+xy^2+yz^2+zx^2 \ge 6xyz$$ Which is obvious by AM-GM