Answer

AM-GM

hsiangshen wrote: Let $a,b,c$ are positive reals. Find the minimum value of $$P = \dfrac{13a+13b+2c}{2a+2b}+\dfrac{24a-b+13c}{2b+2c}+\dfrac{(-a+24b+13c)}{2c+2a}$$. Let $x = 2a+2b,\,y = 2b+2c,\,z = 2c+2a$ then \[2P = \left({\frac {z}{x}}+{\frac {5x}{z}}\right) +\left({\frac {5x}{y}}+{\frac {y}{x}}\right) +\left({\frac {19y}{z}}+ {\frac {19z}{y}}\right) \geqslant 2\sqrt 5 + 2\sqrt 5 + 38 = 2(2 \sqrt 5+19).\]So $P \geqslant 2 \sqrt 5+19,$ equality occur when $(x,\,y,\,z) = \left(\frac{\sqrt{5}t}{5},\,t,\,t\right). $

Let $a,b,c$ are positive reals. Prove that $$\dfrac{13a+13b+2c}{2a+2b}+\dfrac{24a-b+13c}{2b+2c}+\dfrac{-a+24b+13c}{2c+2a}\geq2 \sqrt 5+19.$$ $$\frac{b_0}{a_0}+\frac{c_0}{b_0}=2\sqrt 5$$

Let $a,b,c $ be positive real number . Find the minimum value of $\frac{13a+13b+2c}{a+b}+\frac{24a-b+13c}{b+c}+\frac{-a+24b+13c}{c+a}.$

I just help a math teacher in cram school with this problem So I'd also share my solution here hsiangshen wrote: Let $a,b,c$ be positive reals. Find the minimum value of $$\dfrac{13a+13b+2c}{2a+2b}+\dfrac{24a-b+13c}{2b+2c}+\dfrac{(-a+24b+13c)}{2c+2a}$$. (1)What is the minimum value? (2)If the minimum value occurs when $(a,b,c)=(a_0,b_0,c_0)$,then find $\frac{b_0}{a_0}+\frac{c_0}{b_0}$. Notic that if we plug $(a,b,c)=(ka,kb,kc)$ (where $k$ is an arbritary $\mathbb{R}^+$) in $\dfrac{13a+13b+2c}{2a+2b}+\dfrac{24a-b+13c}{2b+2c}+\dfrac{-a+24b+13c}{2c+2a}$. The value wouldn't change So let's WLOG $a+b=1$ First we fixed $c$ as a constant. $\frac{13a+13b+2c}{2a+2b}+\frac{24a-b+13c}{2b+2c}+\frac{-a+24b+13c}{2c+2a}=\frac{13}2+c+\frac{24-25b+13c}{2c+2b}+\frac{24-25a+13c}{2c+2a}=\frac{13}2+c+f(b)+f(a)$ where $f(x)=\frac{-25x+(13c+24)}{2c+2x}$. It's easy to see that $f$ convex for $x \in [0,+\infty)$ Applying Jensen's Inequality we know $f(a)+f(b) \geq 2f(\frac{a+b}2)=2f(\frac 12)=\frac{26c+23}{2c+1}=13+\frac{10}{2c+1}$ (With equality if and only if $a=b=\frac{1}{2}$) So now we only have to find the minimum of $\frac{13}2+c+13+\frac{10}{2c+1}$ $\frac{13}2+c+13+\frac{10}{2c+1}=19+(c+\frac 12)+\frac{10}{2(c+\frac 12)} \overset{AM-GM}{\geq} 19+2\sqrt{5}$ (With equality if and only if $(c+\frac 12)^2=5 \iff c=-\frac 12\pm\sqrt{5}$ (choose the positive one)) Therefore, the minimun value of $\dfrac{13a+13b+2c}{2a+2b}+\dfrac{24a-b+13c}{2b+2c}+\dfrac{(-a+24b+13c)}{2c+2a}=\boxed{19+2\sqrt{5}}$ With equaly if and only if $(a,b,c)=(\frac{k}{2},\frac{k}{2},-\frac k2+k\sqrt{5})$ (where $k$ is an arbritary $\mathbb{R}^+$) So $\frac{b_0}{a_0}+\frac{c_0}{b_0}=\frac{\frac{k}{2}}{\frac{k}{2}}+\frac{-\frac k2+k\sqrt{5}}{\frac{k}{2}}=\boxed{2\sqrt 5}$