Prove $x^{a+b} + y^{a+b} \le 2$ for $x^a+ y^a = x^b+ y^b$ $(a \ne b)$

Question

Let $a$, $b$ distinct positive numbers and $x$, $y>0$ such that $$x^a+ y^a = x^b+ y^b.$$ Show that $x^{a+b} + y^{a+b} \le 2$.

Notes:

1. It is easy to show that $x^b + y^b\le 2 $. Indeed, assume that $a< b$. We have

$$\left(\frac{x^a + y^a}{2}\right)^{\frac{1}{a}} \le \left(\frac{x^b + y^b}{2}\right)^{\frac{1}{b}}$$ and so

$$\left(\frac{x^a + y^a}{2}\right)^{\frac{1}{a}- \frac{1}{b}}\le 1$$

2. If $x^c+y^c\le 2$ for some $c> 0$ then $x^d+ y^d \le 2$ for all $0\le d \le c$. The argument is similar to the one above.

3. The exponent $a+b$ is the largest possible ( see above).

4. I have seen a problem about $x^2 + y^2 = x^3+ y^3$ on this site.

Any feedback would be welcome!

Answer 1

Remark: This is related to this question.

WLOG, assume that $a < b$.

Fact 1: Let $u, v, r > 0$ be real numbers. Then $$\frac{u^{r+1} + v^{r+1}}{u^r + v^r} \ge \left(\frac{u^{2r+1} + v^{2r+1}}{2}\right)^{\frac{1}{2r+1}}.$$ (The proof is given at the end.)

By Fact 1, letting $r = \frac{a}{b - a}, u = x^{b - a}, v = y^{b - a}$, we have $$\frac{x^{b} + y^{b}}{x^a + y^a} \ge \left(\frac{x^{a + b} + y^{a+b}}{2}\right)^{\frac{b - a}{a + b}}$$ which results in $x^{a + b} + y^{a + b} \le 2$.

We are done.

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Proof of Fact 1:

Since the desired inequality is symmetric and homogeneous, assume that $v = 1$ and $u \ge 1$.

Taking logarithm, it suffices to prove that $$\ln (u^{r+1} + 1) - \ln (u^r + 1) \ge \frac{1}{2r+1}\ln \frac{u^{2r+1} + 1}{2}.$$ Let $f(u) := \mathrm{LHS} - \mathrm{RHS}$. We have $$f'(u) = u^{r-1}\cdot \frac{ru^{2r+2} - (r+1)u^{2r+1} + (r+1)u - r}{(u^{r+1} + 1)(u^r + 1)(u^{2r+1} + 1)}.$$

Let $$g(u) := ru^{2r+2} - (r+1)u^{2r+1} + (r+1)u - r.$$ We have $$g'(u) = r(2r + 2)u^{2r+1} - (r+1)(2r+1)u^{2r} + r + 1$$ and $$g''(u) = 2r(r + 1)(2r + 1)u^{2r - 1}(u - 1).$$

Since $g''(u) \ge 0$ on $u \ge 1$ and $g'(1) = 0$, we have $g'(u) \ge 0$ on $u \ge 1$. Since $g(1) = 0$, we have $g(u) \ge 0$ on $u \ge 1$. Thus, we have $f'(u) \ge 0$ on $u\ge 1$. Since $f(1) = 0$, we have $f(u) \ge 0$ on $u \ge 1$.

We are done.