$x^m+\mu y^m\geq \frac{\min\{1, \mu^{1-m}\}}{2^m} z^m$ where $x+\mu y=z$?

Question

How to prove this:

$$x^m+\mu y^m\geq \frac{\min\{1, \mu^{1-m}\}}{2^m} z^m$$ where $x+\mu y=z$

$m>1$ and $\mu$ is a positive constant

Answer 1

Let $$ a=\frac{x}{x+\mu y}, b=\frac{\mu y}{x+\mu y}. $$ Then $a+b=1$. By using the inequality $(a+b)^m\le 2^{m-1}(a^m+b^m)$ for $a,b>0$, one has \begin{eqnarray} &&\left(\frac{x}{x+\mu y}\right)^m+\mu \left(\frac{y}{x+\mu y}\right)^m\\ &=&a^m+\mu^{1-m}b^m\\ &\ge&\min\{1,\mu^{1-m}\}(a^m+b^m)\\ &\ge&\frac{\min\{1,\mu^{1-m}\}}{2^{m-1}}\\ &\ge&\frac{\min\{1,\mu^{1-m}\}}{2^{m}} \end{eqnarray} and hence $$x^m+\mu y^m\geq \frac{\min\{1, \mu^{1-m}\}}{2^m} (x+\mu y)^m. $$

Answer 2

Let us assume $x, y \ge 0$. This assumption is not superfluous, as the sought-after inequality is false for $x=y=-1$, $\mu=1$, $z = -2$ and $m=3$ since it states in this case that $(-1)^3+(-1)^3 = -2 \ge (-2)^3/2^3 = -1$. More generally, if the inequality holds true for $(x,y, \mu, m)$, then it is false for $(-x, -y, \mu, m)$.

Let us reduce this inequality to the case $\mu=1$, that is, if we assume that this inequality holds true when $\mu=1$, then it holds true for any $\mu > 0$. By this last assumption, we have

$$ x^m + y^m \ge (z/2)^m$$

whenever $x+y=z$ for $x,y \ge 0$. Consequently, for any $\mu > 0$, we have

$$ \mathrm{max}\{1, \mu^{m-1}\}(x^m + \mu y^m) \ge x^m + \mu^m y^m = x^m + (\mu y)^m \ge (z/2)^m$$

whenever $x + \mu y = z$ for $x,y \ge 0$. Therefore,

$$ x^m + \mu y^m \ge \mathrm{min}\{1, \mu^{1-m}\}(z/2)^m $$

which proves the reduction step.

Now here is the proof for $\mu =1$. Note that for $m>1$, the function $f : [0, \infty) \to [0, \infty) : t \mapsto t^m$ is convex. According to Jensen's inequality, we have for any $x,y \ge 0$.

$$ f(\frac{x+y}{2}) \le \frac{f(x) + f(y)}{2} ,\; \mbox{ i.e. } \; 2^{-m}(x+y)^m \le 2^{-1} (x^m+y^m) \le x^m+y^m. $$

Since $z = x+y$, this yields $x^m+y^m \ge (z/2)^m$. In fact, we even got the stronger inequality $x^m+y^m \ge 2(z/2)^m$.