How to prove $\sqrt{\frac{24a+13}{24a+13bc}}+\sqrt{\frac{24b+13}{24b+13ca}}+\sqrt{\frac{24c+13}{24c+13ab}}\ge 3$

Question

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ where $a+b+c=3.$

To prove that: $$\sqrt{\frac{24a+13}{24a+13bc}}+\sqrt{\frac{24b+13}{24b+13ca}}+\sqrt{\frac{24c+13}{24c+13ab}}\ge 3.$$ I've tried to use AM-GM, Cauchy-Schwarz but still can not find an appropriate approach.

The big trouble here is equality occuring at $a=b=c=1$ or $b=c=\dfrac{3}{2};a=0.$

I hope to see a good proof. Thank you!

Answer 1

By Holder $$\left(\sum_{cyc}\sqrt{\frac{24a+13}{24a+13bc}}\right)^2\sum_{cyc}\frac{(24a+13bc)(8a^2+5b^2+5c^2+15ab+15ac+6bc)^3}{24a+13}\geq$$ $$\geq\left(\sum_{cyc}(8a^2+5b^2+5c^2+15ab+15ac+6bc)\right)^3=18^3(a+b+c)^6.$$ Thus, after homogenization it's enough to prove that: $$72(a+b+c)^7\geq\sum_{cyc}\frac{(8a^2+8ab+8ac+13bc)(8a^2+5b^2+5c^2+15ab+15ac+6bc)^3}{85a+13b+13c}$$ or $$\sum_{sym}(171028a^{10}+2490111a^9b+2967068a^8b^2+1600092a^7b^3-4387396a^6b^4-3011931a^5b^5)+$$ $$+abc\sum_{sym}(8266537a^7+12968309a^6b+6428895a^5b^2+2236515a^4b^3)+$$ $$+a^2b^2c^2\sum_{sym}(-2141524a^4-32619909a^3b+9027324a^2b^2+15335781a^2bc)\geq0$$ and sibce by Schur $$\sum_{sym}(a^{10}-2a^9b+a^8bc)\geq0,$$ it's enough to prove that $$\sum_{sym}(8094609a^7+12968309a^6b+6428895a^5b^2+2236515a^4b^3)+$$ $$+abc\sum_{sym}(-21471524a^4-32619909a^3b+9027324a^2b^2+15335781a^2bc)\geq0,$$ which is true by AM-GM and Muirhead.

Answer 2

Some thoughts.

We use the isolated fudging.

It suffices to prove that $$\sqrt{\frac{24a+13}{24a+13bc}} \ge \frac{g(a, b, c)}{h(a, b, c)} \tag{1}$$ where \begin{align*} g(a, b, c) &= 5234a^3 + 102924a^2(b+c) + 105253a(b^2+c^2)\\ &\qquad + 182976abc + 14537(b^3+c^3) + 50591(b^2c+bc^2),\\ h(a, b, c) &= 11436(a^3+b^3+c^3) + 86256(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2)\\ &\qquad + 182976abc. \end{align*}

(1) is true which can be proved by the Buffalo Way (BW).