In a few days, I will be taking part in the final round of my country's maths olympiad and in preparation for that, I was looking through some training exercises.
In some exercises, you had to prove some very symmetric inequalities, such as: $a,b,c>0$. Prove: $$\dfrac{a^3}{b+c} +\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b} \geq \dfrac{a^2+b^2+c^2}{2}$$
I found that these exercises were proven in all kinds of different ways and since you're not allowed to bring a list of formulas to the olympiad, I wanted a more general way to solve these things.
Lemma 1: For a fixed $$S=\sum_{i=1}^{n}x_i$$ the maximum of $$\prod_{i=1}^{n}x_n$$ Lies at $x_1=x_2...=x_n$.
proof: Take the largest and smallest values of the sequence, say $x_k$ and $x_l$. Change both values to $\dfrac{x_k+x_l}{2}$. We see that $S$ remains the same, but with AM-GM $$\big(\dfrac{x_k+x_l}{2}\big)^2 \geq x_kx_l$$, so the product increases. If we keep repeating this proces, we will obtain the desired result.
Lemma 2: For any set of functions, $f_1,f_2...,f_n$ and variables $x_1,x_2, ...,x_m$: $$\min\bigg({\sum_{i=1}^{n}f_i(x_1,x_2...,x_m)}\bigg)$$ Occurs at: $$f_1(x_1,x_2...,x_m)=f_2(x_1,x_2...,x_m)...=f_n(x_1,x_2...,x_m)$$
Proof: Define: $$F(x_1,x_2...,x_m)=\prod_{i=1}^{n}f_i(x_1,x_2...,x_m)$$ The equivalence relation $F(x_1,x_2...,x_m)=F(y_1,y_2...,y_m)$ will partition $\mathbb{R}^m$. By inversing Lemma 1, the lowest sum we can get in an element of the partition, is where $$f_1(x_1,x_2...,x_m)=f_2(x_1,x_2...,x_m)...=f_n(x_1,x_2...,x_m)$$
How to apply my lemma 2
I will now demonstrate how to use lemma 2 by solving an exercise I found somewhere on this site a while ago:
Prove that for $x,y,z>0$, we have: $$\dfrac{x^4}{5x^3+8y^3}+\dfrac{y^4}{5y^3+8z^3}+\dfrac{z^4}{5z^3+8x^3} \geq \dfrac{x+y+z}{13}$$
Subtract the RHS and distribute over terms of the LHS: $$\bigg(\dfrac{x^4}{5x^3+8y^3}-\dfrac{x+y+z}{39}\bigg)+\bigg(\dfrac{y^4}{5y^3+8z^3}-\dfrac{x+y+z}{39}\bigg)+\bigg(\dfrac{z^4}{5z^3+8x^3}-\dfrac{x+y+z}{39}\bigg)\geq0$$ By lemma 2, the LHS reaches it minimum at: $$\dfrac{x^4}{5x^3+8y^3}-\dfrac{x+y+z}{39}=\dfrac{y^4}{5y^3+8z^3}-\dfrac{x+y+z}{39}=\dfrac{z^4}{5z^3+8x^3}-\dfrac{x+y+z}{39}$$, or $x=y=z$. We now find that the minimum is $0$ and we are done.
Questions:
Does this method actually work? (is my proof of the lemmas correct, am I applying them correctly etc.)
If so, why does nobody use this method? (this is why I think my proof is, in fact, incorrect)
Lemma2 is wrong.
Let $f_1(x,y)=x^2$ and $f_2(x,y)=y^2+1$.
Hence, $\min(x^2+y^2+1)=1$ for $x=y=0$, but $f_1(0,0)\neq f_2(0,0)$.