My inequality: For $a,b >0 , a+b=1 $, prove that\[\frac{a^n}{a+b^3}+\frac{b^n}{a^3+b}\ge\frac{2^{4-n}}{5}.(n\in N, n\ge2)\]

wanwan4343 wrote: Let $x,y$ be the positive real numbers with $x+y=1$, and $n$ be the positive integer with $n\ge2$. Prove that \[\frac{x^n}{x+y^3}+\frac{y^n}{x^3+y}\ge\frac{2^{4-n}}{5}\] We need to prove that $5\left(x^{n+3}+y^{n+3}+xy\left(x^{n-1}+y^{n-1}\right)\right)\geq2^{4-n}(x^3y^3+xy+x^4+y^4)$. Since $x^{n+3}+y^{n+3}\geq2^{2-n}(x^5+y^5)$ and $x^{n-1}+y^{n-1}\geq2^{2-n}$, it remains to prove that $5(x^5+y^5+xy)\geq4(x^3y^3+xy+x^4+y^4)$, which is $(1-4xy)(1-4xy+x^2y^2)\geq0$, which is obvious.

Ten years ago, my other inequality ： For $a,b >0 , a+b=1 $, prove that\[\frac{a^n}{a+b^2}+\frac{b^n}{a^2+b}\ge\frac{2^{3-n}}{3}.(n\in N, n\ge2)\]

Dukejukem wrote: We have \[\left(\frac{x^n}{x + y^3} + \frac{y^n}{x^3 + y}\right)\left(x + y^3 + x^3 + y\right) \ge \left(x^{\frac{n}{2}} + y^{\frac{n}{2}}\right)^2 \ge \left(2\left(\frac{x + y}{2}\right)^{\frac{n}{2}}\right)^2 = 2^{2 - n}.\] You proceed to show that $x + y^3 + x^3 + y \ge \frac{5}{4}$ and claim we are done. But you cannot divide inequalities in this way. You would need $x + y^3 + x^3 + y \le \frac{5}{4}$ to conclude, and this unfortunately is not true.

we can use rearrangement inequality to induct on $n$ $$(\frac{x^n}{x+y^3}+\frac{y^n}{x^3+y} ) (x+y) = [\frac{x^{n+1}}{x+y^3}+\frac{y^{n+1}}{x^3+y} ]+ [\frac{x^ny}{x+y^3}+\frac{y^nx}{x^3+y}]$$ the left bracket is greater than the right bracket by rearrangement inequality, so from $$\frac{x^n}{x+y^3}+\frac{y^n}{x^3+y}\ge\frac{2^{4-n}}{5}$$we can get \[\frac{x^{n+1}}{x+y^3}+\frac{y^{n+1}}{x^3+y}\ge\frac{2^{4-(n+1)}}{5}\], now all that's left is to check the base case $n=2$ we proceed by using calculus $$\frac{d}{dx} (\frac{x^2}{x+(1-x)^3}) = \frac{2x({x+(1-x)^3}) - x^2(1-3(1-x)^2)}{(x+(1-x)^3)^2} = \frac{x(1-x)(2-x^2)}{(x+(1-x)^3)^2}$$ if the first term is $f(x)$, the other term is just $f(1-x)$, so its derivative can be found using the chain rule $$\frac{d}{dx} (\frac{(1-x)^2}{1-x+x^3}) = -\frac{x(1-x)(2-(1-x)^2)}{(1-x+x^3)^2}$$ so $$ (2-x^2)(1-x+x^3)^2 = (2-(1-x)^2)(x+(1-x)^3)^2$$ which gives $$ (x-\frac{1}{2})(8x^3(1-x)^3 + 2)=0 $$ so there's only an extremum in $(0,1)$, by evaluating the value at 0.5 and 0.25 we know the extremum is a minimum, plugging in gives the required inequality