2
$\begingroup$

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?

Theorem

$s$ 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, $\vert xy \vert \geq xy$.)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.