Does a computer solution is accept?

Sayan wrote: Let $a,b,c\ge 0$ and $a+b+c=\sqrt2$. Show that \[\frac1{\sqrt{1+a^2}}+\frac1{\sqrt{1+b^2}}+\frac1{\sqrt{1+c^2}} \ge 2+\frac1{\sqrt3}\] By Holder $\left(\sum_{cyc}\frac1{\sqrt{1+a^2}}\right)^2\sum_{cyc}(1+a^2)(a+\sqrt3b+\sqrt3c)^3\geq(1+2\sqrt3)^3(a+b+c)^3$. Hence, it remains to prove that $(1+2\sqrt3)^3(a+b+c)^3\geq\left(2+\frac1{\sqrt3}\right)^2\sum_{cyc}(1+a^2)(a+\sqrt3b+\sqrt3c)^3$. But $(1+2\sqrt3)^3(a+b+c)^3\geq\left(2+\frac1{\sqrt3}\right)^2\sum_{cyc}(1+a^2)(a+\sqrt3b+\sqrt3c)^3\Leftrightarrow$ $\Leftrightarrow3(1+2\sqrt3)(a+b+c)^5\geq\sum_{cyc}((a+b+c)^2+2a^2)(a+\sqrt3b+\sqrt3c)^3\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}(a^4b+a^4c+(3\sqrt3-4)(a^3b^2+a^3c^2)+(15\sqrt3-17)a^3bc+18(\sqrt3-1)a^2b^2c)\geq0$, which is obvious.

Sayan wrote: Let $a,b,c\ge 0$ and $a+b+c=\sqrt2$. Show that \[\frac1{\sqrt{1+a^2}}+\frac1{\sqrt{1+b^2}}+\frac1{\sqrt{1+c^2}} \ge 2+\frac1{\sqrt3}\] In general if $a_1, a_2, \cdots , a_n \ge 0$ and $\sum_{i=1}^n a_i=\sqrt2$ we have \[\sum_{i=1}^n \frac1{\sqrt{1+a_i^2}} \ge (n-1)+\frac1{\sqrt3}\] This inequality follows from two inequalities \[ \dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}} \geq 1+\dfrac{1}{\sqrt{1+(a+b)^2}},\] \[ \dfrac{1}{\sqrt{1+c^2}}+\dfrac{1}{\sqrt{1+(a+b)^2}} \geq 1+\dfrac{1}{\sqrt{3}}.\] The general inequality also folllows from it.

Sayan wrote: Let $a,b,c\ge 0$ and $a+b+c=\sqrt2$. Show that \[\frac1{\sqrt{1+a^2}}+\frac1{\sqrt{1+b^2}}+\frac1{\sqrt{1+c^2}} \ge 2+\frac1{\sqrt3}\]

BMO=Balkan MO or BMO=BrMO?

Sayan wrote: Let $a,b,c\ge 0$ and $a+b+c=\sqrt2$. Show that \[\frac1{\sqrt{1+a^2}}+\frac1{\sqrt{1+b^2}}+\frac1{\sqrt{1+c^2}} \ge 2+\frac1{\sqrt3}\]

http://www.artofproblemsolving.com/community/c6h1210775p6000747: Let $a,b$ and $c$ be positive real numbers such that $a+b+c=abc$. Prove that $$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \le \frac{3}{2}$$Let $a,b$ and $c$ be positive real numbers such that $a+b+c=abc$.Prove that $$\frac{2}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \le \frac{9}{4}$$Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \le \frac{3}{\sqrt{2}}$$Let $a,b$ and $c$ be positive real numbers such that $a+b+c=\sqrt{3}$. Prove that $$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \leq \frac{3\sqrt{3}}{2}$$Let $a,b$ and $c$ be positive real numbers such that $ a^2 + b^2 + c^2+abc = 4$. Prove that $$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}}\ge \frac{3}{\sqrt{2}}$$

Hi,everyone. This problem is BALKAN MO 2012 SL problem $(A2)$,not BALKAN MO 2012

Proof that for all positive real number $x,y,z$ such that $xyz=\frac{27}{64}$ Then $\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}} \leq \frac{12}{5}$

bdtilove123 wrote: Does a computer solution is accept? give it a shot!

The initial problem was proposed by me for the Balkan MO 2012 and it was shortlisted.

quykhtn-qa1 wrote: \[ \dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}} \geq 1+\dfrac{1}{\sqrt{1+(a+b)^2}},\] Let $a,b\ge 0$, and $a+b \le \sqrt{2}$. Show that

edit: wrong sol

levifb wrote: Let's prove something stronger. Define $f(x) = \frac{1}{\sqrt{1+x^2}}$. $f'(x) = -\frac{1}{2}(x^2 + 1)^{-\frac{3}{2}} \cdot 2x \Rightarrow f''(x) = \frac{3}{2}(x^2+1)^{-\frac{5}{2}} \cdot 2x \geq 0 ~ \forall ~x \in \mathbb{R}_{\geq 0}$ Since the second derivative of $f$ is greater or equal than zero, we can use the Jensen's Inequality: $\frac{f(a) + f(b) + f(c)}{3} \geq f(\frac{a+b+c}{3}) = f(\frac{\sqrt{2}}{3}) = \frac{3}{\sqrt{11}}$ $\Rightarrow \frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \geq \frac{9}{\sqrt{11}} \approx 2,736 \geq 2,57735 \approx 2 + \frac{1}{\sqrt{3}}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \blacksquare$

Kkkkkk derivei errado fds

Can this be someway solved with Jensen?

There have been a lot of algebra mistakes in this thread. Jensen is not applicable because the function $\frac{1}{\sqrt{1+x^2}}$ is concave down on $0\le x< \frac{1}{\sqrt{2}}$ and concave up on $x>\frac{1}{\sqrt{2}}$. Since $a,b,c\in [0, \sqrt{2}]$, there is exactly one point of inflection on the interval. This makes this inequality a candidate for n-1 EV and of course still the tangent line trick.

Skravin wrote: bdtilove123 wrote: Does a computer solution is accept? give it a shot! WLOG, assume that $a \ge b \ge c$. We have $$\frac{1}{\sqrt{1 + b^2}} + \frac{1}{\sqrt{1 + c^2}} \ge \frac{1}{1 + b^2/2} + \frac{1}{1 + c^2/2}.$$We have \begin{align*} &\frac{1}{1 + b^2/2} + \frac{1}{1 + c^2/2} - 1 - \frac{1}{1 + (b + c)^2/2}\\[6pt] ={}&\frac{bc(-{b}^{3}c-2\,{b}^{2}{c}^{2}-b{c}^{3}-4\,bc+8)}{ \left( {c}^{2}+2 \right) \left( {b}^{2}+2 \right) \left( {b}^{2}+2 \,bc+{c}^{2}+2 \right) }\\ \ge{}& 0 \end{align*}where we use $b, c < 1$. It suffices to prove that $$\frac{1}{\sqrt{1 + a^2}} + 1 + \frac{1}{1 + (b + c)^2/2} \ge 2 + \frac{1}{\sqrt 3}$$or $$\frac{1}{\sqrt{1 + a^2}} + 1 + \frac{1}{1 + (\sqrt 2 - a)^2/2} \ge 2 + \frac{1}{\sqrt 3}$$or $$\frac{1}{\sqrt{1 + a^2}} - \frac{1}{\sqrt 3} \ge 1 - \frac{1}{1 + (\sqrt 2 - a)^2/2}$$or $$\frac{2 - a^2}{\sqrt{1 + a^2}\sqrt 3 (\sqrt 3 + \sqrt{1 + a^2})} \ge \frac{(\sqrt 2 - a)^2}{2 + (\sqrt 2 - a)^2}$$or $$\frac{\sqrt 2 + a}{\sqrt 3\, (a^2 + 1) + 3\sqrt{1 + a^2}} \ge \frac{\sqrt 2 - a}{2 + (\sqrt 2 - a)^2}$$or $$\frac{\sqrt 2 + a}{2 (a^2 + 1) + 3(1 + a^2/2)} \ge \frac{\sqrt 2 - a}{2 + (\sqrt 2 - a)^2}$$or (letting $a = \sqrt 2\, x$ for $1/3 \le x \le 1$) $$\frac{\sqrt 2\, (1 + x)}{2 (2x^2 + 1) + 3(1 + x^2)} \ge \frac{\sqrt 2\, (1 - x)}{2 + 2(1 - x)^2}$$which is true. We are done.

Let $f(a,b,c)=\sum_{cyc}(1+a^2)^{-\frac{1}{2}}, g(a,b,c)=a+b+c-\sqrt{2}\Longrightarrow L=f(a,b,c)-\lambda g(a,b,c)$ $\frac{\partial L}{\partial a}=-\frac{a}{(\sqrt{1+a^2})^3}-\lambda=0$ $\frac{\partial L}{\partial b}=-\frac{b}{(\sqrt{1+b^2})^3}-\lambda=0$ $\frac{\partial L}{\partial c}=-\frac{c}{(\sqrt{1+c^2})^3}-\lambda=0$ Thus: $\frac{a}{(\sqrt{1+a^2})^3}=\frac{b}{(\sqrt{1+b^2})^3}=\frac{c}{(\sqrt{1+c^2})^3}$ (1)(2)(3) respectively. From (1) and (2) we get: $\frac{a}{(\sqrt{1+a^2})^3}=\frac{b}{(\sqrt{1+ b^2)^3}}\Longrightarrow \frac{a^2}{1+a^6+3a^4+3a^2}=\frac{b^2}{1+b^6+3b^4+3b^2}$, furthermore $a^2-b^2+3a^2b^4-3a^4b^2=a^6b^2-a^2b^6\Longrightarrow a^2b^2(b^2-a^2)(a^2+b^2+3)=b^2-a^2$, thus either $a=b$ or $a^2b^2(a^2+b^2+3)=1$ however the second can never hold since $a^2b^2$ is bounded by $\frac{1}{3}$, and from the equations (1) and (3). (We procced identically for $c$) Thus $a=b=c\Longrightarrow a=b=c=\frac{\sqrt{2}}{3}$, furthermore minimum is achieved when $f(\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3})=\frac{9}{\sqrt{11}}>2+\frac{1}{\sqrt{3}}$