Is $\frac1{\sqrt{(a^2+b^2)(c^2+b^2)}}+\frac1{\sqrt{(b^2+c^2)(c^2+a^2)}}+\frac1{\sqrt{(a^2+b^2)(a^2+c^2)}}\le\frac{(a+b+c)}{2abc}$?

Question

Is $\frac1{\sqrt{(a^2+b^2)(c^2+b^2)}}+\frac1{\sqrt{(b^2+c^2)(c^2+a^2)}}+\frac1{\sqrt{(a^2+b^2)(a^2+c^2)}}\le\frac{(a+b+c)}{2abc}$?

I think yes, by seeing the arithmetic geometric means inequality, but am unable to exactly apply it. Applying the inequality gives an inequality involving a cube root, but isnt any cube root on the RHS. Any help. Thanks beforehand.

PS: This is problem S392 in Mathematical Reflections.

Answer 1

hint: True. Apply AM-GM inequality for the denominators: example: $\sqrt{(x^2+y^2)(y^2+z^2)} \ge \sqrt{(2xy)(2yz)} = 2y\sqrt{xz}$, and rationalize denominators, then next use Cauchy-Schwarz inequality as follows: $\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{z}+\sqrt{z}\sqrt{x} \le x+y+z$. Can you proceed ?

alternatively, you can prove it by Cauchy-Schwarz inequality, followed by AM-GM as before: $AB+BC+CA \le A^2+B^2+C^2$, and $a^2+b^2 \ge 2ab$, for $A = \dfrac{1}{\sqrt{a^2+b^2}}, B = \dfrac{1}{\sqrt{b^2+c^2}}$, etc..Can you also proceed with this way?