Let $x,y$ be positive real numbers such that $x\neq y$ and $3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value: $M=\frac{4}{x^{2}}+\frac{4}{y^{2}}+\frac{1}{(x-y)^{2}}$.
I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $\frac{ax}{y}$ while $x,y$ are some constants.
Help me to solve this problem.
Let $x=2$ and $y=1$.
Thus, we'll get a value $6$.
We'll prove that it's a minimal value.
Indeed, let $x=2a$ and $y=b$.
Thus, $3a^2+2b^3=5$ and we need to prove that $$\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{(2a-b)^2}\geq6.$$ But by AM-GM $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2\geq$$ $$\geq3a^2+3b^2-1+3\sqrt[3]{(b^3)^2\cdot1}-3b^2=3(a^2+b^2)-1,$$ which gives $2\geq a^2+b^2.$
Id est, $$\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{(2a-b)^2}\geq\frac{1}{2}(a^2+b^2)\left(\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{(2a-b)^2}\right).$$ Now, let $a=tb$.
Thus, we need to prove that $$(t^2+1)\left(\frac{1}{t^2}+4+\frac{1}{(2t-1)^2}\right)\geq12$$ or $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)\geq0,$$ which is obviously true for $t\geq1.$
But for $0<t\leq1$ we obtain: $$16t^4+16t^3-7t^2-2t+1=$$ $$=9t^3(1-t)+\left(5t^2+\frac{7}{10}t-\frac{5}{6}\right)^2+\frac{1}{900}(759t^2-750t+275)\geq0.$$ Done!
Also, we can use derivative here.
Since $$(t^2+1)\left(\frac{1}{t^2}+4+\frac{1}{(2t-1)^2}\right)\geq12$$ it's $f(t)\geq0,$ where $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$ we obtain $t_{min}=\frac{1}{3}$ or $t_{min}=1$ and since $f\left(\frac{1}{3}\right)>f(1)=0$, we are done.