Proving $\sum_{cyc}\frac{a}{\sqrt{8a+bc}}\le \frac{\sqrt{a+b+c+abc}}{2}$ if $ab+bc+ca=3.$

Question

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=3.$ Prove that$$\color{black}{\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\le \frac{\sqrt{a+b+c+abc}}{2}.}$$ Source: Vo Quoc Ba Can.

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This symmetrical inequality is extremally tight! It was in a test of Vo Quoc Ba Can's class. (Toan hoc muon mau).

Some thoughts.

A big trouble here is equality occuring $(a,b,c)=(1,1,1)$ or $(a,b,c)=(0,\sqrt{3},\sqrt{3})$ and its permutations.

I'm afraid of that the author has a nice proof as usual like AM-GM, C-S,...etc. Unfortunately, I don't see any bright ideas.

I think the Bacteria method helps for this situation well but I failed to find an appropriate using.

For example, we can trivially use Cauchy-Schwarz inequality as$$\color{black}{\sum_{cyc}\frac{a}{\sqrt{8a+bc}}=\sum_{cyc}\sqrt{\frac{a}{8a+bc}}\cdot\sqrt{a}\le \sqrt{(a+b+c)\sum_{cyc}\frac{a}{8a+bc}}\le \frac{\sqrt{a+b+c+abc}}{2}}$$which the last inequality saves the equality occuring but is wrong at $a=b=\dfrac{1}{2};c=\dfrac{11}{4}.$

Also, the equality cases $a=b=c; a=b;c=0$ recalls us about Schur inequality.

Indeed, by using Schur of third degree $$abc\ge \frac{(a+b+c)[4(ab+bc+ca)-(a+b+c)^2]}{9}=\frac{(a+b+c)[12-(a+b+c)^2]}{9},$$ it implies $$RHS\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$ (I just mention in case $21-(a+b+c)^2\ge 0.$)

Now, I saw an interesting fact that $$\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$ It means that the original inequality is more stronger than Schur$$RHS\ge LHS\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$ How hard could it be!

I hope to see some ideas for this nice problem. Thank you for your interest.

Answer 1

Some thoughts.

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\left(\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\right)^2\\[6pt] \le{}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\left(\sum_{\mathrm{cyc}} (3a + abc)\right)\\[6pt] ={}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\cdot 3(a + b + c + abc). \end{align*}

It suffices to prove that $$\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )} \le \frac{1}{12}. \tag{1}$$

(1) is true which is verified by Mathematica.

Answer 2

By C-S $$\sum_{cyc}\frac{a}{\sqrt{8a+bc}}\leq\sqrt{\sum_{cyc}\frac{a}{a+2}\sum_{cyc}\frac{a(a+2)}{8a+bc}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a}{a+2}\sum_{cyc}\frac{a(a+2)}{8a+bc}\leq\frac{a+b+c+abc}{4}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$ and $abc=w^3$.

Thus, $v=1$ and we need to prove that: $$\sum_{cyc}\frac{a}{a+2v}\sum_{cyc}\frac{a(a+2v)}{8va+bc}\leq\frac{3uv^2+w^3}{4v^3}$$ or $$\tfrac{3w^3+12v^3+12uv^2}{w^3+14v^3+12uv^2}\cdot\tfrac{216uv^5+288v^6+27u^3w^3-126u^2vw^3-27uv^2w^3+384v^3w^3+3w^6}{w^6+464v^3w^3+72u^2vw^3-384uv^2w^3+576v^6}\leq\frac{3uv^2+w^3}{4v^3}$$ or $f(u)\geq0,$ where $$f(u)=(3uv^2+w^3)(w^3+14v^3+12uv^2)(w^6+464v^3w^3+72u^2vw^3-384uv^2w^3+576v^6)-$$ $$-36v^3(w^3+4v^3+4uv^2)(72uv^5+96v^6-42u^2vw^3-9uv^2w^3+116v^3w^3+9u^3w^3+w^6).$$ But $$f'(u)=3v(6912uv^9+1728u^3v^4w^3-6048u^2v^5w^3+5280uv^6w^3+3376v^7w^3+$$ $$+756u^2v^2w^6-2136uv^3w^6+602v^4w^6+48uw^9-123vw^9)\geq0$$ because by Maclaurin $u\geq v\geq w$ and also $v^4\geq uw^3$.

Thus, $f$ increases and it remains to prove that $f(u)\geq0$ for a minimal value of $u$, which by $uvw$ happens for equality case of two variables.

Let $b=a$.

Thus, $c=\frac{3-a^2}{2a}$, where $0<a\leq\sqrt3$ and we need to prove that: $$(a-1)^2(3-a^2)(a^8-20a^7+46a^6+204a^5-459a^4-534a^3+780a^2+882a+108)\geq0,$$ which is true for $0<a\leq\sqrt3$.

It's interesting that $$\sum_{cyc}\frac{a}{a+1}\sum_{cyc}\frac{a(a+1)}{8a+bc}\leq\frac{a+b+c+abc}{4}$$ is wrong, but for $b=a$ and $c=\frac{3-a^2}{2a}$ it's true!