# Prove that $a\sqrt{b^2+8c^2}+b\sqrt{c^2+8a^2}+c\sqrt{a^2+8b^2}\leqslant(a+b+c)^2$ for all positive real numbers $a,b,c$

Let $$a,b,c>0$$ , Prove that $$a\sqrt{b^2+8c^2}+b\sqrt{c^2+8a^2}+c\sqrt{a^2+8b^2}\leqslant(a+b+c)^2$$

I've tried to get rid of the square roots by substitution and Cauchy-Schwarz, but I didn't find any substitution which doesn't leave any square roots, and my attempts using Cauchy-Schwarz are always too crude.

How to prove this inequality?

• WLOG we can assume that $a \ge b \ge c$. I wasn't able to prove the full formula, only for the case when $a \ge \frac{7+\sqrt{45}}{2}b \approx 6.85b$. It goes like this: $$a\sqrt{b^2+8c^2}+b\sqrt{c^2+8a^2}+c\sqrt{a^2+8b^2} \le a\sqrt{b^2+8b^2}+b\sqrt{a^2+8a^2}+c\sqrt{a^2+8a^2} = a\sqrt{9b^2}+b\sqrt{9a^2}+b\sqrt{9a^2} = 3ab+6ab = 9ab \stackrel{a \ge 8.85b}{\le} (a+b)^2 < (a+b+c)^2$$ We likely need a less rough estimate than to replace some $b$ with $a$, and some $c$ with $b$ to be able to prove this. Commented Mar 24, 2023 at 17:11
• See for some proofs: artofproblemsolving.com/community/c6t243f6h2707806_sqrta28b2 Commented Mar 25, 2023 at 0:34

## 2 Answers

This is too long for a comment and I don't see it being rigorous, but I will share my thoughts on this.

First note that dividing both sides of inequality by $$a+b+c$$ we have $$\frac{a}{a+b+c}\sqrt{b^2 + 8 c^2} + \frac{b}{a+b+c}\sqrt{c^2 + 8 a^2} + \frac{c}{a+b+c}\sqrt{a^2 + 8 b^2} \le a + b + c.$$ Thus the LHS is a weighted sum of a function $$f(x,y) = \sqrt{x^2 + 8y^2}$$. Jensen's inequality (I think...) allows us to write $$f\left(\sum w_i \vec x_i \right) \ge \sum w_i f(\vec x_i)$$ where the $$w_i$$ are the weights above and $$\vec x_i$$ are the arguments to be inserted into the function $$f$$. Thus after simplifying $$LHS \le f(z,z) = 3 z$$ where $$z = \frac{ab + bc + ac}{a+b+c}$$. Finally, note that $$(a+b+c)^2 - 3(ab + bc + ac) = a^2 + b^2 + c^2 - ab - ac - bc \ge a^2 + b^2 + c^2 - 2ab - 2ac - 2bc = (a-b-c)^2 \ge 0$$ Implies that $$\frac{3(ab+bc+ac)}{a +b + c} \le a+b+c.$$ This completes the proof.

EDIT: To calculate the sum (note I am writing the arguments in $$f$$ as the tuple $$(x,y)$$ in what follows): $$\frac{a(b,c) + b(c,a) + c(a,b)}{a+b+c} = \left(\frac{ab + bc + ac}{a + b + c}, \frac{ab + bc + ac}{a + b + c} \right) = (z,z)$$ Feeding this into the function $$f(z,z) = 3z$$.

• (You're missing a summation sign in $f( w_i \vec{x_i})$.) How did you calculate that $\sum w_i x_i = \frac{ab+bc+ca}{a+b+c}$? Commented Mar 24, 2023 at 18:27
• Edited - let me know if anymore questions. But won't be able to say much in regards to rigor here. Commented Mar 24, 2023 at 18:36
• @Gregory $LHS \le \frac{3(ab+bc+ac)}{a +b + c}$ is not true. Check $a = b = 2, c = 1$. Commented Mar 24, 2023 at 23:22
• @Gregory Also, $f(x, y)$ is convex, so it should be $f\left(\sum w_i \vec x_i \right) \le \sum w_i f(\vec x_i)$. Commented Mar 24, 2023 at 23:25

Because by C-S $$\sum_{cyc}a\sqrt{b^2+8c^2}\leq\sqrt{\sum_{cyc}\frac{a(b^2+8c^2)}{b+2c}\sum_{cyc}a(b+2c)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a(b^2+8c^2)}{b+2c}\leq\frac{(a+b+c)^4}{3(ab+ac+bc)}$$ or $$\sum_{cyc}(2a^6b+4a^6c+12a^5b^2+18a^5c^2-2a^4b^3-28a^4c^3)+$$ $$+abc\sum_{cyc}(33a^4+18a^3b-4a^3c-39a^2b^2-12a^2bc)\geq0,$$ which is true because by AM-GM $$\sum_{cyc}(2a^6b+4a^6c+12a^5b^2+18a^5c^2-2a^4b^3-28a^4c^3)+$$ $$+abc\sum_{cyc}(33a^4+18a^3b-6a^3c-39a^2b^2-12a^2bc)=$$ $$=\sum_{cyc}\left(2a^6b+4a^6c+\frac{28}{3}a^5b^2+\frac{44}{3}a^5c^2-2a^4b^3-28a^4c^3\right)+$$ $$+\sum_{cyc}\left(\frac{8}{3}a^5b^2+\frac{10}{3}a^5c^2\right)+$$ $$+abc\sum_{cyc}(33a^4+18a^3b-6a^3c-39a^2b^2-12a^2bc)=$$ $$=\frac{2}{3}\sum_{cyc}(a-b)^2ab(3a^3+20a^2b+34ab^2+6b^3)+$$ $$+\sum_{cyc}\left(\frac{8}{3}a^5b^2+\frac{10}{3}a^5c^2\right)+$$ $$+abc\sum_{cyc}(33a^4+18a^3b-6a^3c-39a^2b^2-12a^2bc)\geq$$ $$\geq abc\sum_{cyc}(33a^4+18a^3b-6a^3c-33a^2b^2-12a^2bc)\geq$$ $$\geq abc\sum_{cyc}(27a^4+6a^3b-33a^2b^2)=3abc\sum_{cyc}a^2(a-b)(9a+11b)=$$ $$=3abc\sum_{cyc}(a^2(a-b)(9a+11b)-5(a^4-b^4))=3abc\sum_{cyc}(a-b)^2(4a^2+10ab+5b^2)\geq0.$$