Conjugates and Young's Inequality

Example: Let $0<r<2$, $p=\frac{2}{r}$, and $q=\frac{2}{2-r}$. Then the previous result, applied to $x^r$ and $y^{2-r}$, becomes $$x^ry^{2-r}\leq\frac{r}{2}x^2+\frac{2-r}{2}y^2,$$ which provides us with a quadratic majorization for $x^r$ for all $0<r<2.$ We have equality if and only if $x=y$.

showMe <- function (b, r, up = 5) {
    a <- seq (0, up, length = 100)
    ar <- a ^ r
    plot (a, ar, type = "l", col = "RED",
          ylab="f(a)", lwd = 2)
    br <- (r * (a ^ 2)) / 2 + (((2 - r)  * (b ^ 2))/ 2)
    br <- br / (b ^ (2 - r))
    lines (a, br, col = "GREEN", lwd = 2)
    abline (v = b, lwd = 2)
}

Here is an example with $y=2$ and $r=1.5$.

And an example with $y=2$ and $r=0.5$.

One important application of these results is majorization of powers of Euclidean distances $(\sqrt{x'Ax})^r$ by quadratic forms. We find, for $0<r<2$, $$(\sqrt{x'Ax})^r\leq\frac{1} {(\sqrt{y'Ay})^{(2-r)}}\left\{\frac{r}{2}x'Ax+\frac{2-r}{2}y'Ay\right\}$$