Let $a,b,c>0.$ Prove that $$ \sqrt{9a^2+(a+b+c)^2}+\sqrt{9b^2+(a+b+c)^2}+\sqrt{9c^2+(a+b+c)^2} \ge \sqrt{12(a^2+b^2+c^2)+14(a+b+c)^2}$$
I see it on Facebook here. I tried Mincopxki, but it doesn't help. $$\sum\limits_{cyc} \sqrt{9a^2+(a+b+c)^2} \ge \sqrt{9(a+b+c)^2+9(a+b+c)^2}=\sqrt{18}(a+b+c).$$ But the following $a^2+b^2+c^2 \le ab+bc+ca$ is not true.
I think it's very hard, I also tried AM-GM, Cauchy - Schwarz, but it is not word too. Please help me, thank all.
On
Another way.
We need to prove that: $$\sum_{cyc}\sqrt{10a^2+b^2+c^2+2(ab+ac+bc)}\geq\sqrt{\sum_{cyc}(26a^2+28abc)}.$$
Let $a\geq b\geq c$.
Thus, $$\sqrt{a^2+10b^2+c^2+2(ab+ac+bc)}+\sqrt{a^2+b^2+10c^2+2(ab+ac+bc)}\geq$$ $$\geq\sqrt{4a^2+17.5b^2+17.5c^2+8ab+8ac+17bc}$$ it's $$4\sqrt{(a^2+10b^2+c^2+2(ab+ac+bc))(a^2+b^2+10c^2+2(ab+ac+bc)}\geq$$ $$\geq4a^2+13b^2+13c^2+8ab+8ac+26bc$$ or $$(b-c)^2(8a^2+16ab+16ac-b^2-38bc-c^2)\geq0,$$ which is true for $a\geq b\geq c$.
Thus, it remains to prove that:
$$\sqrt{4a^2+17.5b^2+17.5c^2+8ab+8ac+17bc}+\sqrt{10a^2+b^2+c^2+2(ab+ac+bc)}\geq$$$$\geq\sqrt{\sum\limits_{cyc}(26a^2+28bc)}$$ or $$8(8a^2+35b^2+35c^2+16ab+16ac+34bc)(10a^2+b^2+c^2+2(ab+ac+bc))\geq$$$$\geq9(8a^2+5b^2+5c^2+12ab+12ac+6bc)^2$$ or $$64a^4-320(b+c)a^3+48(23b^2-2bc+23c^2)a^2-8(b+c)(49b^2+62bc+49c^2)a+$$ $$+55b^4+292b^3c+330b^2c^2+292bc^3+55c^4\geq0$$ or $$64a^4-320(b+c)a^3+48(23(b+c)^2-48bc)a^2-8(b+c)(49(b+c)^2-36bc)a+$$ $$+55(b+c)^4+72bc((b+c)^2-2bc).$$ Now, let $b+c=constant$ and $bc=x$.
Thus, we need to prove that $f(x)\geq0,$ where $$f(x)=64a^4-320(b+c)a^3+48(23(b+c)^2-48x)a^2-8(b+c)(49(b+c)^2-36x)a+$$ $$+55(b+c)^4+72x((b+c)^2-2x).$$ But $$f'(x)=-48^2a^2+288(b+c)a+72(b+c)^2-288x=$$ $$=72(-32a^2+4(b+c)a+(b+c)^2-4bc)\leq$$ $$\leq72(-32a^2+4(3-a)a+(3-a)^2)=72(9+6a-35a^2)\leq0$$ because $a\geq1$.
Id est, $f$ decreases and it's enough to prove $f(x)\geq0$ for a maximal value of $bc$, which happens for $b=c$.
Thus, we need to prove that: $$64a^4-640a^3b+2112a^2b^2-2560ab^3+1024b^4\geq0$$ or $$64(a-b)^2(a-4b)^2\geq0$$ and we are done!
On
There is a nice proof.
Due to homogeneity, let $a+b+c=3$. The inequality becomes: \begin{align*} \sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1} \ge \sqrt{\frac{4}{3}\left(a^2+b^2+c^2 \right)+14} \end{align*} Squaring both sizes gives: \begin{align*} \sum_{\text{cyc}}{\sqrt{\left(a^2+1 \right) \left(b^2+1 \right)}} \ge \frac{1}{6} \left(a^2+b^2+c^2 \right) +\frac{11}{2}\end{align*} or, \begin{align*} \sum_{\text{cyc}} \left[ \sqrt{\left(a^2+1 \right) \left(b^2+1 \right)} -a-b\right] \ge \frac{a^2+b^2+c^2 -3}{6}\end{align*} or, \begin{align*} \sum_{\text{cyc}} \frac{\left(ab-1 \right)^2}{\sqrt{\left(a^2+1 \right) \left(b^2+1 \right)}+a+b}\ge \frac{a^2+b^2+c^2 -3}{6}\end{align*} Applying AM-GM inequality: \begin{align*} \sqrt{\left(a^2+1 \right) \left(b^2+1 \right)} +a+b &\le \frac{a^2+b^2+2}{2} +a+b \\ &= \frac{\left(a+1 \right)^2 + \left(b+1 \right)^2}{2}\end{align*} It suffices to prove that: \begin{align*} \sum_{\text{cyc}} \frac{\left(ab-1 \right)^2}{\left(a+1 \right)^2 + \left(b+1 \right)^2} \ge \frac{a^2+b^2+c^2-3}{12}\end{align*} WLOG, assume $a\ge b \ge c$ then $a \ge 1$. Using Cauchy-Schwarz inequality, we have: \begin{align*} \frac{\left(ab-1 \right)^2}{\left(a+1 \right)^2 + \left(b+1 \right)^2} + \frac{\left(ca-1 \right)^2}{\left(c+1 \right)^2 + \left(a+1 \right)^2} & \ge \frac{\left(ab-ca \right)^2}{2 \left(a+1 \right)^2 + \left(b+1 \right)^2 + \left(c+1 \right)^2} \\ & \ge \frac{\left(ab-ca \right)^2}{2 \left(a+a \right)^2 + \left(a+a \right)^2 + \left(a+a \right)^2} \\ & = \frac{1}{16} \left(b-c \right)^2\end{align*} Also, it is easy to see that: \begin{align*} \frac{\left(bc-1 \right)^2}{\left(b+1 \right)^2 + \left(c+1 \right)^2} \ge \frac{1}{2} \left(\frac{b+c}{2} - 1\right)^2 \Longleftrightarrow \left( b-c\right)^2 \left( b^2+4bc+c^2-2b-2c-2\right) \le 0\end{align*} Which is true because: \begin{align} \left(b^2+2bc+c^2\right) + 2bc \le \left(b+c\right)^2 + \frac{\left(b+c\right)^2}{2} \le 2 \left(b+c \right) + 2 \end{align} where $b+c \le 2$.
Thus, \begin{align} \sum_{\text{cyc}} \frac{\left(ab-1 \right)^2}{\left(a+1 \right)^2 + \left(b+1 \right)^2} & \ge \frac{1}{16}\left(b-c \right)^2 + \frac{1}{2} \left(\frac{b+c}{2} - 1 \right)^2 \\ & = \frac{1}{16}\left(b-c \right)^2 +\frac{1}{72} \left(b+c-2a \right)^2 \\ & \ge \frac{1}{24} \left(b-c\right)^2 + \frac{1}{72} \left(b+c-2a \right)^2 = \frac{a^2+b^2+c^2-3}{12}\end{align} We have Q.E.D.
On
Using calculus, there is a relatively simple solution and the results are more general. When $a,b,c \ge 0$, $a+b+c \ge 0$ and $3(a+b+c)^2 \ge 3(a^2+b^2+c^2) \ge (a+b+c)^2$. We look for minimum and maximum of LHS under the constraints $\{a+b+c=3p,\ a^2+b^2+c^2=3p^2+6r^2\}$, with $p \ge r \ge 0$. Below we allow $a,b,c$ to take negative values too, but the constraints imply that $ab+bc+ca \ge 0$.
Viewing $a, b, c$ as coordinates in 3D Euclidean space, the set of points satisfying the constraints is a circle, which is a closed set without boundary, so both minimum and maximum of LHS must exist and they are at stationary points of the function. Introduce Lagrange multipliers $3m, 3n/2$ for the constraints, take derivatives, the equations of the stationary condition are
$$\frac{a}{\sqrt{a^2+p^2}}+m+na=0 \tag{1}\label{eq1}$$ $$\frac{b}{\sqrt{b^2+p^2}}+m+nb=0 \tag{2}\label{eq2}$$ $$\frac{c}{\sqrt{c^2+p^2}}+m+nc=0 \tag{3}\label{eq3}$$
So $a,b,c$ are zeros of the function $f(x)=\frac{x}{\sqrt{x^2+p^2}}+m+nx$. Assume $a \le b \le c$. We prove that $a,b,c$ cannot be all distinct. If they are, the first derivative of $f(x)$ must have at least one zero in each of $(a,b)$ and $(b,c)$, and the second derivative must have a zero in $(a,c)$. Since $f''(x)=-3p^2x(x^2+p^2)^{-5/2}$ only has one zero at $x=0$, we must have $a<0<c$. Combined with $a+b+c\ge0$ and $ab+bc+ca \ge 0$, we must have $b>0, (a+c)>0$. $\eqref{eq2}c - \eqref{eq3}b => m<0$, and $\eqref{eq1}c-\eqref{eq3}a => \sqrt{a^2+p^2}>\sqrt{c^2+p^2} => |a|>|c|$. But this contradicts with $a+c>0$.
Given symmetry between $a,b,c$, we can assume $b=c$. Together with the two constraints, $a,b,c$ for the stationary points can be solved: $a=p \pm 2r, b=c=p \mp r$, and $$ LHS=3(\sqrt{(p \pm 2r)^2+p^2} + 2\sqrt{(p \mp r)^2+p^2}\ ) $$ Plugging in a value like $r=p$ shows that the top sign corresponds to minimum and the bottom sign corresponds to maximum. The minimum is a tighter bound than RHS. To see that, apply Minkowski: $$ 3(\sqrt{(p+2r)^2+p^2} + \sqrt{(2p)^2 + (2p-2r)^2}\ ) \ge 3(\sqrt{(3p+2r)^2+(3p-2r)^2}\ )=3(\sqrt{18p^2+8r^2}\ )=RHS $$
Let $a+b+c=3$ and $a\geq b\geq c$.
Thus, we need to prove that: $$3\sum_{cyc}\sqrt{a^2+1}\geq\sqrt{\sum_{cyc}(26a^2+28ab)}$$ or $$9\sum_{cyc}\left(a^2+1+2\sqrt{(a^2+1)(b^2+1)}\right)\geq\sum_{cyc}(26a^2+28ab)$$ or $$18\sum_{cyc}\sqrt{(a^2+1)(b^2+1)}\geq\sum_{cyc}(26a^2+28ab-9a^2-3a^2-6ab)$$ or $$\sum_{cyc}(-14a^2-22ab+9(a^2+1+b^2+1))\geq9\sum_{cyc}\left(a^2+1-2\sqrt{(a^2+1)(b^2+1)}+b^2+1\right)$$ or $$\sum_{cyc}(4a^2-22ab+6a^2+12ab)\geq9\sum_{cyc}\left(\sqrt{a^2+1}-\sqrt{b^2+1}\right)^2$$ or $$\sum_{cyc}(10a^2-10ab)\geq9\sum_{cyc}\frac{(a^2-b^2)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}$$ or $$\sum_{cyc}(a-b)^2\left(5-\frac{9(a+b)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}\right)\geq0.$$ Now, by Minkowski(triangle inequality): $$5-\frac{9(b+c)^2}{\left(\sqrt{b^2+1}+\sqrt{c^2+1}\right)^2}\geq5-\frac{9(b+c)^2}{(b+c)^2+4}=\frac{20-4(b+c)^2}{(b+c)^2+4}\geq\frac{20-4\cdot2^2}{(b+c)^2+4}>0.$$ Also, by Minkowski again $$5-\frac{9(a+c)^2}{\left(\sqrt{a^2+1}+\sqrt{c^2+1}\right)^2}=5-\frac{9(a+c)^2}{\left(\sqrt{a^2+1}+\sqrt{1+c^2}\right)^2}\geq$$ $$\geq5-\frac{9(a+c)^2}{(a+1)^2+(c+1)^2}=\frac{2(5a+5c+5-2a^2-9ac-2c^2)}{(a+1)^2+(c+1)^2}=$$ $$=\frac{2(15(a+c)(a+b+c)+5(a+b+c)^2-9(2a^2+9ac+2c^2))}{9((a+1)^2+(c+1)^2)}\geq$$ $$\geq\frac{2(15(a+c)(a+2c)+5(a+2c)^2-9(2a^2+9ac+2c^2))}{9((a+1)^2+(c+1)^2)}=\frac{4(a-4c)^2}{9((a+1)^2+(c+1)^2)}\geq0.$$ Thus, by Minkowski again we obtain: $$\sum_{cyc}(a-b)^2\left(5-\frac{9(a+b)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}\right)\geq$$ $$\geq(a-b)^2\left(5-\frac{9(a+b)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}\right)+(a-c)^2\left(5-\frac{9(a+c)^2}{\left(\sqrt{a^2+1}+\sqrt{c^2+1}\right)^2}\right)\geq$$ $$\geq(a-b)^2\left(5-\frac{9(a+b)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}\right)+(a-b)^2\left(5-\frac{9(a+c)^2}{\left(\sqrt{a^2+1}+\sqrt{c^2+1}\right)^2}\right)=$$ $$=(a-b)^2\left(10-\frac{9(a+b)^2}{\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)^2}-\frac{9(a+c)^2}{\left(\sqrt{a^2+1}+\sqrt{c^2+1}\right)^2}\right)\geq$$ $$\geq(a-b)^2\left(10-\frac{9(a+b)^2}{(a+1)^2+(b+1)^2}-\frac{9(a+c)^2}{(a+1)^2+(c+1)^2}\right)$$ and it's enough to prove that: $$10-\frac{9(a+b)^2}{\left(a+\frac{a+b+c}{3}\right)^2+\left(b+\frac{a+b+c}{3}\right)^2}-\frac{9(a+c)^2}{\left(a+\frac{a+b+c}{3}\right)^2+\left(c+\frac{a+b+c}{3}\right)^2}\geq0$$ or $$68a^4-220(b+c)a^3+(147b^2+78bc+147c^2)a^2+(b+c)(443b^2-518bc+443c^2)a+$$ $$+89b^4+545b^3c+588b^2c^2+545bc^3+89c^4\geq0$$ or $$68a^4-220(b+c)a^3+(147(b+c)^2-216bc)a^2+(b+c)(443(b+c)^2-1404bc)a+$$ $$+89(b+c)^4+27bc(7(b+c)^2-10bc)\geq0.$$ Now, let $b+c=constant$ and $bc=x$.
Thus, we need to prove that $f(x)\geq0,$ where $$f(x)=68a^4-220(b+c)a^3+(147(b+c)^2-216x)a^2+(b+c)(443(b+c)^2-1404x)a+$$ $$+89(b+c)^4+27x(7(b+c)^2-10x).$$ But $$f'(x)=-216a^2-1404(b+c)a+189(b+c)^2-540x<0,$$ which says that it's enough to prove $f(x)\geq0$ for a maximal value of $x$, which happens for $b=c$.
Now, let $b=c=1$.
Thus, it remains to prove that: $$68a^4-440a^3+372a^2+736a+1856\geq0$$ or $$(a-4)^2(17a^2+26a+29)\geq0$$ and we are done.