# Proving $\sqrt{5a^2+4ab}+\sqrt{5b^2+4bc}+\sqrt{5c^2+4ca}\ge 9$ when $ab+bc+ca=3.$

Problem. Let $$a,b,c\ge 0: ab+bc+ca=3.$$ Prove that $$\sqrt{5a^2+4ab}+\sqrt{5b^2+4bc}+\sqrt{5c^2+4ca}\ge 9.$$

A big trouble is around $$(a,b,c)=(2,3/2,0).$$

I've tried to square both side and we'll prove $$5(a^2+b^2+c^2)+4(ab+bc+ca)+2\sum_{cyc}\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\ge 81,$$ or $$5(a^2+b^2+c^2)+2\sum_{cyc}\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\ge 69.$$ We have $$a^2+b^2+c^2\ge ab+bc+ca=3,$$and it's enough to prove $$\sum_{cyc}\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\ge 27.$$ By using CBS inequality $$\sum_{cyc}\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\ge \sum_{cyc}(5ab+4b\sqrt{ca})=15+4\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right).$$ The rest is proving $$\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge 3$$but $$3=ab+bc+ca\ge \sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right).$$

EDIT: There is some problem in the original proof below. Specifically $$p^3r\geqslant q^3$$ does not necessarily hold. I found this inequality accidently at that time but did not check the correctness. So, sorry for my carelessness.

Here is a new proof (still complicated, though).

Our goal is to prove $$\sum\sqrt{5a^2+4ab}\geqslant9=3\sqrt{3(ab+bc+ca)}.$$ We can prove $$\sum\sqrt{\frac{5a^2+4ab}{3(ab+bc+ca)}}\geqslant3.$$

We can simplify the square root operation, by the following equality: $$\sqrt{t}\geqslant\frac{6t^3+20t^2+6t}{t^3+15t^2+15t+1},$$ which can be derived from $$(\sqrt t-1)^6\geqslant0\iff t^3+15t^2+15t+1\geqslant\sqrt t(6t^2+20t+6)$$.

Therefore $$\sqrt t-1\geqslant\frac{(t - 1) (5 t^2 + 10 t + 1)}{(t + 1) (t^2 + 14 t + 1)}=:f(t).$$

Let $$u=\frac{5a^2+4bc}{3(ab+bc+ca)},\, v=\frac{5b^2+4bc}{3(ab+bc+ca)},\, w=\frac{5c^2+4ca}{3(ab+bc+ca)}.$$ It can be proved $$f(u)+f(v)+f(w)\geqslant0.$$

It can be verified with the help of Mathematica. We can assume without loss of generality $$c=\min(a,b,c)$$. We need to consider both the case $$c\leqslant a\leqslant b$$ and $$c\leqslant b\leqslant a$$. The first case is simplier than the second. For the second case, the minimum is achieved only if $$c=0$$, so by the homogeneity we just need to consider $$(a,b,c)=(a,1,0)$$. The rest is just to verify the positiveness for all $$a$$ in such case.

The proof is done.

PS: If the estimation of $$\sqrt t$$ is based on $$(\sqrt t-1)^4\geqslant0$$ instead of $$(\sqrt t-1)^6$$, it seems the positiveness cannot be guaranteed in the case $$c\leqslant b\leqslant a$$.

This is NOT a nice solution so only the idea is sketched.

You have proved $$\sum\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant 15+4\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$$. Hence $$\sum\sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant15+12(abc)^{2/3}.$$ Then, in order to prove the original inequality, we just need to prove $$\begin{gathered} \vphantom{\sum}\\ \impliedby\vphantom{a^2} \\ \impliedby\vphantom{a^2} \end{gathered} \begin{gathered} 5(a^2+b^2+c^2)+2\sum \sqrt{5a^2+4ab}\cdot\sqrt{5b^2+4bc}\geqslant 69\\ 5(a^2+b^2+c^2)+24(abc)^{2/3}\geqslant39\\ 5(a^2+b^2+c^2)+24(abc)^{2/3}\geqslant13(ab+bc+ca). \end{gathered}$$

This is true but in a short time I haven't found an elegant proof.

There is an ugly proof available. Let $$a=x^3$$, $$b=y^3$$, $$c=z^3$$, $$p=x+y+z$$, $$q=xy+yz+xz$$, $$r=xyz$$. We can prove $$5\sum x^6+24(xyz)^2\geqslant 13\sum x^3y^3,$$ using the facts that $$\begin{gathered} \sum x^6= p^6−6p^4q+6p^3r+9p^2q^2−2q^3−12pqr+3r^2,\\ \sum x^3y^3=q^3-3pqr+3r^2,\\ p^3r\geqslant q^3,\\ p^3r\geqslant 3pqr. \end{gathered}$$ After simplification, the only thing left is $$5p^6+45p^2q^2\geqslant5p^2\cdot 6p^2q,$$ which is obvious.

• I check $a=2,b=3/2,c=0.$ Am I missing something?
– user1272006
Commented Dec 24, 2023 at 9:13
• @FurryNick You're right. I did not check the source in which the inequality $p^3r\geqslant q^3$ was given (just found it accidently). It turns out it's not reliable. I will give another proof later. Commented Dec 24, 2023 at 22:31
• I recommend that you avoid using "EDIT:". Instead, revise the answer so it reads well for someone who encounters it for the first time. I suggest deleting the old, incorrect answer. The old version can still be found in the revision history. We're looking to build an archive of knowledge that will be useful to others in the future.
– D.W.
Commented Dec 25, 2023 at 4:46