Minimizing $ \sum\limits_{cyc}\frac{\sqrt{5a+8bc}}{8a+5bc}$ with $ \sum\limits_{cyc}ab=1$

Question

Olympiad inequality. Let $a,b,c\ge 0: ab+bc+ca=1.$ Find the minimal value $P$ of $$f:=\frac{\sqrt{5a+8bc}}{8a+5bc}+\frac{\sqrt{5b+8ca}}{8b+5ca}+\frac{\sqrt{5c+8ab}}{8c+5ab}.$$ Note: Often Stack Exchange asks to show some work before answering the question. This inequality was used as a proposed problem for the National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They didn't believe that any student can solve this problem in the $3$ hour timeframe.

It seems that minimum is achieved at $(a,b,c)=(0,1,1).$ I've tried to prove $$f\ge \frac{\sqrt{5}}{4}+\frac{2\sqrt{2}}{5}. \tag{1}$$ A big problem around here is $a=b=c=\dfrac{\sqrt{3}}{3}$ since $LHS_{(1)}-RHS_{(1)}\approx 0.000151$

I hope to see some ideas. Thank you!

Update 1: There are two answers and RiverLi's proof is good but not easy to full it by hand.

Update 2: You can see also here.

Answer 1

Some thoughts.

Let $$x := 5a + 8bc, \quad y := 5b + 8ca, \quad z := 5c + 8ab,$$ $$u := 8a + 5bc, \quad v := 8b + 5ca, \quad w := 8c + 5ab.$$ Let $$A := \frac{2\sqrt{2} - \sqrt{5}}{3}, \quad B := \frac{8\sqrt{5} - 10\sqrt{2}}{3}.$$

By AM-GM, we have $$ \frac{\sqrt{5a+8bc}}{8a+5bc} = \frac{x}{u}\cdot \frac{2}{2\sqrt{x}} \ge \frac{x}{u} \cdot \frac{2}{\frac{x}{Ax + B} + (Ax + B)} = \frac{2x(Ax + B)}{ux + u(Ax + B)^2}. $$

It suffices to prove that $$\frac{2x(Ax + B)}{ux + u(Ax + B)^2} + \frac{2y(Ay + B)}{vy + v(Ay + B)^2} + \frac{2z(Az + B)}{wz + w(Az + B)^2} \ge \frac{\sqrt{5}}{4}+\frac{2\sqrt{2}}{5}. \tag{1}$$

We use the pqr method. Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$. We have $$p^2 \ge 3q, \quad q^2 \ge 3pr, \quad r \ge \frac{4pq - p^3}{9}. \tag{2}$$ (Note: $r \ge \frac{4pq - p^3}{9}$ is the degree three Schur inequality.)

(1) is equivalently written as $$C(p, r) \sqrt{5} + D(p, r)\sqrt{2} \ge 0 \tag{3}$$ where $C(p, r)$ and $D(p, r)$ are two polynomials. ¹⁾

(3) is true for all $p, r \ge 0$ with $1 \ge 3pr$ and $p^2 \ge 3$ and $r \ge \frac{4p - p^3}{9}$, which is verified by Mathematica - a Computer Algebra System (CAS). Thus, (1) is true. We need to find a proof of (3) which can be verified easily.

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Footnotes.

¹⁾

\begin{align*} C &:= -307200000\,{p}^{6}{r}^{3}-313344000\,{p}^{5}{r}^{3}-1175040000\,{p}^{ 4}{r}^{4}\\ &\qquad -3138240000\,{p}^{5}{r}^{2}+10636876800\,{p}^{4}{r}^{3}+ 560793600\,{p}^{3}{r}^{4}\\ &\qquad -1400832000\,{p}^{2}{r}^{5}+3206832000\,{p}^{ 4}{r}^{2}-30808175040\,{p}^{3}{r}^{3}\\ &\qquad +15838519680\,{p}^{2}{r}^{4}+ 1699430400\,p{r}^{5}-491520000\,{r}^{6}\\ &\qquad +2202240000\,{p}^{4}r+ 92830480200\,{p}^{3}{r}^{2}-282601800\,{p}^{2}{r}^{3}\\ &\qquad -29734119264\,p{r }^{4}+1220244480\,{r}^{5}+10496784000\,{p}^{3}r\\ &\qquad -195792425400\,{p}^{2}{ r}^{2}+171792599925\,p{r}^{3}-1397911875\,{r}^{4}\\ &\qquad -33643248000\,{p}^{2} r-32289450120\,p{r}^{2}-179729064750\,{r}^{3}\\ &\qquad +768000000\,{p}^{2}+ 40110748800\,pr+173550992400\,{r}^{2}\\ &\qquad +4915200000\,p-31985030400\,r- 12902400000 \end{align*} and \begin{align*} D &:= -491520000\,{p}^{6}{r}^{3}-319488000\,{p}^{5}{r}^{3}-1880064000\,{p}^{ 4}{r}^{4}\\ &\qquad -5061120000\,{p}^{5}{r}^{2}+15576883200\,{p}^{4}{r}^{3}+ 1421721600\,{p}^{3}{r}^{4}\\ &\qquad -2241331200\,{p}^{2}{r}^{5}+6120960000\,{p}^ {4}{r}^{2}-46081374720\,{p}^{3}{r}^{3}\\ &\qquad +22093286400\,{p}^{2}{r}^{4}+ 3092643840\,p{r}^{5}-786432000\,{r}^{6}\\ &\qquad +3494400000\,{p}^{4}r+ 143923780800\,{p}^{3}{r}^{2}+694726080\,{p}^{2}{r}^{3}\\ &\qquad -43404609792\,p{ r}^{4}+331849728\,{r}^{5}+16226880000\,{p}^{3}r\\ &\qquad -307929041040\,{p}^{2}{ r}^{2}+266002645320\,p{r}^{3}+157445640\,{r}^{4}\\ &\qquad -51839472000\,{p}^{2}r -50083230240\,p{r}^{2}-284401827024\,{r}^{3}\\ &\qquad +1248000000\,{p}^{2}+ 62068857600\,pr+275071716480\,{r}^{2}\\ &\qquad +7641600000\,p-50523340800\,r- 20275200000. \end{align*}

Answer 2

Not an solution, just some thoughts

Let $f(a,b,c)$ be the expression we want to minimize. Suppose one of the variables is $0$, and the other two are inverses of each other. WLOG, assume $a=0, b=\frac{1}{c}$. We get:

$f(0,\frac{1}{c},c) = \frac{\sqrt{8}}{5} + \frac{\sqrt{\frac{5}{c}}}{\frac{8}{c}} + \frac{\sqrt{5c}}{8c} = \frac{\sqrt{8}}{5} + \frac{\sqrt{5}}{8}\left(\sqrt{c} + \frac{1}{\sqrt{c}}\right)$

By AM-GM, it's now clear $f(0,\frac{1}{c},c) \geq \frac{2\sqrt{2}}{5} + \frac{\sqrt{5}}{4}$.

From this point on, assume, WLOG, $a\geq b\geq c>0$.

Claim 1: $\min(a,b,c) \leq 1$.

Proof: Suppose not. Then $ac+ab+bc=1$ is impossible. $\blacksquare$

Claim 2: If $a\geq b\geq c$, then $1 \geq b\geq c$.

Proof: Suppose not. Then $a\geq b > 1$. Meaning $ab>1$. Meaning $ab+bc+ac=1$ is impossible. $\blacksquare$

Claim 3: $abc \leq \frac{1}{3\sqrt{3}}$.

Proof: AM-GM gives $\frac{ab+bc+ac}{3} \geq \sqrt[3]{a^{2}b^{2}c^{2}} \implies \frac{1}{3} \geq \sqrt[3]{a^{2}b^{2}c^{2}} \implies \frac{1}{3\sqrt{3}} \geq abc$ $\blacksquare$

Claim 4: $a+b+c \geq \sqrt[3]{3\sqrt{3}}$

Proof: Direct application of claim 3 and AM-GM $\blacksquare$

Claim 5: $a+b+c \geq \sqrt{3}$

Let $S=a+b+c$ and $P = abc$. We can say that $a,b,c$ are the roots of an polynomial $g(x) = x^{3} - Sx^{2} + x - P$. For $g$ to have 3 real roots, we must have that $g$ has an critical point above $0$, and another below $0$. Meaning

$3x^{2} -2Sx + 1 = 0 \implies x = \frac{2S\pm2\sqrt{S^{2} - 3}}{2} = S \pm \sqrt{S^{2}-3}$

Must have real roots, so $S \geq \sqrt{3}$.$\blacksquare$

Claim 6: If $b=c$, then $a\geq \frac{1}{\sqrt{3}}$, and $b=c=\sqrt{a^{2}-1}-a$

Suppose we know that two of the variables are equal. Then, either $a=b=1$ (which we already solved), or $1 \geq b=c$. In this situation we have that $ac + ab + bc = 1 \implies b(b+2a)=1 \implies b^{2}+2ab - 1=0 \implies b=\frac{-2a\pm 2\sqrt{a^{2}+1}}{2} = \sqrt{a^{2}+1}-a$ (since $b>0$).

Enforcing $a\geq b$, we get that this has an solution whenever $a\geq \frac{1}{\sqrt{3}}$. $\blacksquare$

So, if we plug in $f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a)$, $a\geq \frac1{\sqrt{3}}$, we fairly easily check $f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a) > f(1,1,0)$

Meaning if there is an minimum, and it is not $f(1,1,0)$, it must occur with $a\neq b\neq c$. So it suffices to show this is not possible.

From this point on, we suppose $a>b>c$. We also know $1\geq b > c$.

I ran out of ideas here. I suspect one can prove that $f(a,b,c) \geq f(a,\sqrt{a^{2}+1}-a,\sqrt{a^{2}+1}-a)$ whenever $a\geq \frac{1}{\sqrt{3}}$, which would be enough, but I have been unable to do so.