Lukaluce wrote: Let $a, b, c$ be positive real numbers such that $a^2 + b^2 + c^2 = \frac{1}{4}.$ Prove that $$\frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}} + \frac{1}{\sqrt{a^2 + b^2}} \le \frac{\sqrt{2}}{(a + b)(b + c)(c + a)}.$$ Solution of Zhangyanzong:

Attachments:

$\sum{\frac{1}{\sqrt{b^2+c^2}}} \le \sum{\frac{\sqrt{2}}{b+c}}$ simply $\sum{(a+b)(b+c)} \le 1$ (?). Which is equal to $a^2+b^2+c^2+3ab+3bc+3ca \le 1(?)$ ,as, $ab+bc+ca \le \frac{1}{4}=a^2+b^2+c^2$. Which is clear.

Lukaluce wrote: Let $a, b, c$ be positive real numbers such that $$a^2 + b^2 + c^2 = \frac{1}{4}.$$ Prove that $$\frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}} + \frac{1}{\sqrt{a^2 + b^2}} \le \frac{\sqrt{2}}{(a + b)(b + c)(c + a)}.$$ $(1+1)(b^2+c^2)\geqslant (b+c)^2\Leftrightarrow \frac{1}{\sqrt{b^2+c^2}}\leq \frac{\sqrt{2}}{b+c}$ So we need to prove that: $\sum \frac{1}{a+b}\leqslant \frac{1}{(a+b)(b+c)(c+a)}\Leftrightarrow \sum (a+b)(a+c)\leqslant 1\Leftrightarrow a^2+b^2+c^2+3(ab+bc+ca)\leqslant 1$ Which is true from the condition of the problem

The key is to notice that $$\frac{(b+c)^2}{2}\leq b^2+c^2\Rightarrow \frac{1}{\sqrt{b^2+c^2}}\leq \frac{\sqrt{2}}{(b+c)}$$Now to finish notice that $$\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}+\frac{1}{\sqrt{a^2+b^2}}\leq \frac{\sqrt{2}}{(b+c)}+\frac{\sqrt{2}}{(c+a)}+\frac{\sqrt{2}}{(a+b)}=$$$$\frac{\sqrt{2}((a^2+b^2+c^2)+3(ab+bc+ca))}{(a+b)(b+c)(c+a)}\leq \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}$$

Denote $x=\sqrt{a^2+b^2}, y=\sqrt{b^2+c^2}, z=\sqrt{c^2+a^2}$. Then the left-hand side is \[\frac{xy+yz+zx}{xyz}\leq \frac{x^2+y^2+z^2}{xyz} = \frac{2(a^2+b^2+c^2)}{xyz}=\frac{1}{2xyz}.\]We now wish to prove that $\frac{1}{2xyz}\leq \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}$, which follows from multiplying $2x^2=2(a^2+b^2)\geq (a+b)^2$ and its cyclic forms.

an extension here: Generalization 1 Suppose $a_1,a_2,\cdots,a_n$ ($n\geq 2$) are positive reals satisfying $$\sum_{cyc}{a_1^{n-1}}=\dfrac{1}{\left(n-1\right)^{n-1}}$$ Then prove that $$\sum_{cyc}{\dfrac{1}{\sqrt{a_1+a_2+\cdots+a_{n-1}}}}\leq \dfrac{\sqrt{n}}{\prod\limits_{cyc}{\left(a_1+a_2+\cdots+a_{n-1}\right)}}$$

By AM-QM, we have $\sqrt{\tfrac{a^2+b^2}{2}} \ge \tfrac{a+b}{2}$, so it follows that $\tfrac{1}{\sqrt{a^2+b^2}} \le \tfrac{\sqrt{2}}{a+b}$. Cyclicly summing, we have \[\sum_{cyc} \frac{1}{\sqrt{a^2+b^2}} \le \sum_{cyc} \frac{\sqrt{2}}{a+b}.\]Thus, it suffices to show that \[\sum_{cyc} \frac{1}{a+b} \le \frac{1}{(a+b)(b+c)(b+c)}.\]Clearing the fractions and expanding, this is equivalent to $a^2+b^2+c^2 + 3(ab+bc+ca) \le 1$. However, since $ab+bc+ca\le a^2+b^2+c^2$, we have $a^2+b^2+c^2 + 3(ab+bc+ca) \le 4(a^2+b^2+c^2) = 1$, so we are done. $\blacksquare$

Lukaluce wrote: Let $a, b, c$ be positive real numbers such that $$a^2 + b^2 + c^2 = \frac{1}{4}.$$ Prove that $$\frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}} + \frac{1}{\sqrt{a^2 + b^2}} \le \frac{\sqrt{2}}{(a + b)(b + c)(c + a)}.$$ Proposed by Petar Filipovski, Macedonia $$\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\sum_{cyc}\frac{\sqrt{2}}{a+b}=\frac{\sqrt{2}((a^2+b^2+c^2)+3(ab+bc+ca))}{(a+b)(b+c)(c+a)}\leq \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}$$

Here is the solution for my pupils aged 13, 14 and 15 those who have not learnt Cauchy-Schwarz inequality, Jensen's inequality and the Quadratic Mean yet. Using the $AM-GM$ inequality and $2(x^2+y^2)=(x+y)^2+(x-y)^2\geq (x+y)^2$ for all real numbers $x$ and $y$, we have $\frac1{\sqrt{b^2 + c^2}} + \frac1{\sqrt{c^2 + a^2}} + \frac1{\sqrt{a^2 + b^2}}$ $=\frac{\sqrt{(c^2+a^2)(a^2+b^2)}+\sqrt{(a^2+b^2)(b^2+c^2)}+\sqrt{(b^2+c^2)(c^2+a^2)}}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$ $\leq \frac{\dfrac{c^2+a^2+a^2+b^2}2+\dfrac{a^2+b^2+b^2+c^2}2+\dfrac{b^2+c^2+c^2+a^2}2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$ $=\frac{2(a^2+b^2+c^2)}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}\Longleftarrow {a^2+b^2+c^2=\frac14}$ $=\frac1{2\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}=\frac{\sqrt{2}}{\sqrt{2(a^2+b^2)}\sqrt{2(b^2+c^2)}\sqrt{2(c^2+a^2)}}$ $\leq \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}\ \because a, b, c>0.$ Equality attains if and only if $a^2+b^2+c^2=\frac14$ and $a=b=c>0\Longleftrightarrow a=b=c=\frac1{2\sqrt{3}}.$ Kunihiko Chikaya/July 1, 2024

Firstly, use $x^2+y^2 \geq \frac{(x+y)^2}{2}$ to get: $$\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{c+b} \leq \frac{1}{(a+b)(b+c)(a+c)}$$Multiply by the denominator to get: $$a^2+b^2+c^2+3(ab+bc+ac) \leq 1$$But $ab+bc+ac \leq a^2+b^2+c^2 \leq \frac{1}{4}$, from which the desired inequality follows

Dear AoPS admins and moderators, may I know why my last post regarding the name "Macedonia" in this thread was deleted without any notice from your side? I find it utterly shameful that your actions included the deletion of my post, and not the fix in the name of the country which proposed this problem in a prestigious international contest.

We use the following well-known inequalities: $$\sqrt{\frac{a^2 + b^2}{2}}\ge \frac{a+b}{2} \implies \frac{1}{\sqrt{a^2 + b^2}} \le \frac{\sqrt{2}}{a+b}$$(QM-AM), $$a^2 + b^2 + c^2 \ge ab + bc + ca$$(Consequence of AM-GM). Note that we have $$LHS = \sum_{cyc} \frac{1}{\sqrt{b^2 + c^2}} \stackrel{\rm QM-AM}{\le} \sqrt{2} \times \sum_{cyc} \frac{1}{a+b}$$$$ = \frac{\sqrt{2}}{\prod_{cyc} (a+b)} \times \left(\sum_{cyc} (a+b)(a+c)\right) = \frac{\sqrt{2}}{\prod_{cyc} (a+b)} \times\left(\sum_{cyc} a^2 + 3\sum_{cyc}ab\right)$$$$\stackrel{\rm Consequence of AM-GM}{\le} \frac{\sqrt{2}}{\prod_{cyc} (a+b)} \times\left(4\sum_{cyc} a^2 \right) = \frac{\sqrt{2}}{\prod_{cyc}(a+b)} = RHS.$$$\square$

By AMQM reciprocated, we have $$\sqrt{\frac{2}{b^2+c^2}} \leq \frac{2}{b+c}$$and likewise for the others. The desired inequality then turns into $$\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b} \leq \frac{1}{(a+b)(b+c)(a+c)} \implies a^2+b^2+c^2+3(ab+bc+ac) \leq 1 \implies ab+bc+ac \leq \frac14$$We also have $a+b+c \leq \frac{\sqrt{3}}{2}$ by AMQM again, so the desired conclusion follows.