Prove $\frac{x^{n}}{x+y^3}+\frac{y^{n}}{y+x^3} \geqslant \frac{2^{4-n}}{5}$ for $x, y > 0$ with $x+y=1$

Question

Positive real numbers $x$,$y$ satisfy $x+y=1$. For any integer $n \geqslant 2$, show that $$\frac{x^{n}}{x+y^3}+\frac{y^{n}}{y+x^3} \geqslant \frac{2^{4-n}}{5}.$$

So far, I've tried some elementary ways, such as AM-GM, Holder, induction. Maybe certain method mentioned above is feasible, but I've made little progress, please help.

Answer 1

The hint:

For all $n\geq3$ prove that $f(x)=\frac{x^n}{x+(1-x)^3}$ is a convex function and use Jensen.

Consider the case $n=2$.

Indeed, for $n=2$ we need to prove that $$\frac{x^2}{x+(1-x)^3}+\frac{(1-x)^2}{1-x+x^3}\geq\frac{4}{5}$$ or $$(2x-1)^2(x^4-2x^3+5x^2-4x+1)\geq0,$$ which is obvious.

Let $n\geq3$ and $f(x)=\frac{x^n}{x+(1-x)^2}.$

Thus, $$f''(x)=\tfrac{x^{n-2}((x+(1-x)^3)^2n^2-(x+(1-x)^3)(1-6x+15x^2-7x^3)n+2x^2(6x^4-24x^3+33x^2-15x+1))}{(x+(1-x)^3)^3}.$$ For $n=3$ we have $f''(x)>0$ because in this case by AM-GM: $$(x+(1-x)^3)^2n^2-(x+(1-x)^3)(1-6x+15x^2-7x^3)n+$$ $$+2x^2(6x^4-24x^3+33x^2-15x+1)=$$ $$=9(x+(1-x)^3)^2-3(x+(1-x)^3)(1-6x+15x^2-7x^3)+$$ $$+2x^2(6x^4-24x^3+33x^2-15x+1)=$$ $$=2(3x^5-6x^4+6x^3+x^2-6x+3)\geq2(3x^3+x^2+1+1+1-6x)\geq$$ $$\geq2\left(5\sqrt[5]{3x^3\cdot x^2\cdot1^3}-6x\right)=2x\left(5\sqrt[5]3-6\right)>0.$$ Now, $$3(x+(1-x)^3)-(1-6x+15x^2-7x^3)=2(x-1)^2(2x+1)>0.$$ Thus, for all $n\geq4$ we obtain: $$(x+(1-x)^3)^2n^2-(x+(1-x)^3)(1-6x+15x^2-7x^3)n+$$ $$+2x^2(6x^4-24x^3+33x^2-15x+1)=$$ $$=n(x+(1-x)^3)(3(x+(1-x)^3)-(1-6x+15x^2-7x^3))+(n^2-3n)(x+(1-x)^3)^2+$$ $$+2x^2(6x^4-24x^3+33x^2-15x+1)\geq$$ $$\geq(n^2-3n)(x+(1-x)^3)^2+2x^2(6x^4-24x^3+33x^2-15x+1)\geq$$ $$\geq4(x+(1-x)^3)^2+2x^2(6x^4-24x^3+33x^2-15x+1)=$$ $$=2(8x^6-36x^5+59x^4-43x^3+21x^2-8x+2)\geq$$ $$\geq2(8x^6-36x^5+59x^4-43x^3+13x^2)=2x^2(8x^4-36x^3+59x^2-43x+13)=$$ $$=2x^2\left(2\left(2x^2-\frac{9}{2}x+2\right)^2+\frac{1}{2}(5x^2-14x+10)\right)>0.$$

Answer 2

The desired inequality is written as $$\frac{(2x)^n}{x + y^3} + \frac{(2y)^n}{y + x^3} \ge \frac{16}{5}. \tag{1}$$

If $n = 2, 3$, (1) is directly verified.

In the following, assume that $n \ge 4$.

By Cauchy-Bunyakovsky-Schwarz inequality, it suffices to prove that $$\frac{\left((2x)^{n/2} + (2y)^{n/2}\right)^2}{x + y^3 + y + x^3} \ge \frac{16}{5}. \tag{2}$$

By the power mean inequality, we have $$\left(\frac{(2x)^{n/2} + (2y)^{n/2}}{2}\right)^{2/n} \ge \left(\frac{(2x)^2 + (2y)^2}{2}\right)^{1/2}$$ which results in $$(2x)^{n/2} + (2y)^{n/2}\ge 2 \left(\frac{(2x)^2 + (2y)^2}{2}\right)^{n/4}. \tag{3}$$

Using $\frac{(2x)^2 + (2y)^2}{2} = 2(x^2 + y^2) \ge (x + y)^2 = 1$ and $n/4 \ge 1$, we have $$\left(\frac{(2x)^2 + (2y)^2}{2}\right)^{n/4} \ge \frac{(2x)^2 + (2y)^2}{2}. \tag{4}$$

From (3) and (4), we have $$(2x)^{n/2} + (2y)^{n/2} \ge 4x^2 + 4y^2. \tag{5}$$

From (2) and (5), it suffices to prove that $$\frac{(4x^2 + 4y^2)^2}{x + y^3 + y + x^3} \ge \frac{16}{5}$$ or $$\frac{(80x^2 - 80x + 48)(1 - 2x)^2}{15x^2 - 15x + 10} \ge 0$$ which is true.

We are done.