Is there an argument for the following inequality using only the basic algebraic properties of inequalities and using the Cauchy-Schwartz Inequality, Young's Inequality, the Arithmetic-Geometric Mean Inequality, and Bernouli's Inequality?


$s$ and $t$ are conjugate (positive) real numbers~---~they total to 1~--~and $x$ and $y$ are any real numbers. \begin{equation*} sx^{k} + ty^{k} \geq \sum_{i=0}^{k} \binom{k}{i} (sx)^{i}(ty)^{k-i} \end{equation*} for any nonnegative integer $k$.

Demonstration in the case $k=2$

According to the Arithmetic-Geometric Mean Inequality, for any positive real numbers $x$ and $y$, \begin{equation*} \frac{x^{2} + y^{2}}{2} \geq xy . \end{equation*} So, \begin{equation*} st(x^{2} + y^{2}) \geq 2stxy , \end{equation*} \begin{equation*} t(sx^{2}) + s(ty^{2}) \geq 2stxy , \end{equation*} \begin{equation*} (1 - s)sx^{2} + (1 - t)ty^{2} \geq 2stxy , \end{equation*} \begin{equation*} sx^{2} + ty^{2} \geq s^{2}x^{2} + 2stxy + t^{2}y^{2} . \end{equation*} (If either $x$ or $y$ is a negative real number, $\vert xy \vert \geq xy$.)


Using the binomial formula, your inequality becomes $$ (sx + ty)^k \le s x^k + t y^k \, . $$ With $f(x) = x^k$ and using $s+t =1$ this becomes $$ f(sx + (1-s)y) \le s f(x) + (1-s)f(y) \quad \text{ for } 0 \le s \le 1 $$ which is exactly the condition for $f$ to be convex.

If $k$ is even then $f(x) = x^k$ is convex on $\Bbb R$, so that the inequality is true for any $x, y \in \Bbb R$.

If $k$ is odd then $f(x) = x^k$ is convex on $[0, \infty)$, so that the inequality is true for any $x, y \ge 0$.

But the inequality is false for $x, y < 0$ and odd $k$ because then $z \to z^k$ is concave on the negative real numbers so that the inverse inequality holds.

  • $\begingroup$ Yes, I see that you recognized the identity $\sum_{i=0}^{k} \binom{k}{i} (sx)^{i} (ty)^{k-i} = (sx + ty)^{k}$, and you cite the convexity of the power function $x^{k}$ on the interval $[0, \, \infty)$. $\endgroup$ – user74973 Jan 10 '16 at 15:10
  • $\begingroup$ @user74973: Are you asking why $x^k$ is convex? The second derivative is $\ge 0$. $\endgroup$ – Martin R Jan 10 '16 at 15:16
  • $\begingroup$ I was just commenting on your argument. $\endgroup$ – user74973 Jan 10 '16 at 15:21
  • $\begingroup$ @user74973: If anything is missing or unclear in my argument then please let me know. $\endgroup$ – Martin R Jan 10 '16 at 15:25

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