# An Inequality with Constraint IX

### Problem

Leo Giugiuc has kindly posted a folklore problem at the CutTheKnotMath facebook page, along with a solution (Proof 1) by Leo Giugiuc and Dan Sitaru. Proof 2 is by Kunihiko Chikaya. The illustration is by Gary Davis. ### Proof 1

The function $f(t)\;$ is convex on $(0,\infty ),\;$ hence, by Jensen's inequality, $\displaystyle\frac{x^3+y^3}{2}\ge\left(\frac{x+y}{2}\right)^3,\;$ implying $x+y\le 2.$

Let $1-x=a\;$ and $1-y=b.\;$ Then $a^3+b^3=(a+b)(a^2-ab+b^2).\;$ But $a^2-ab+b^2\ge 0\;$ and $a+b=2-(x+y)\ge 0,\;$ so that $a^3+b^3\ge 0,\;$ or, $(1-x)^3+(1-y)^3\ge 0.\;$ Equivalently,

$2-(x^3+y^3)+3(x^2+y^2)-3(x+y)\ge 0,$

which is exactly $x^2+y^2\ge x+y.$

### Proof 2

$\displaystyle x^3+y^3=\frac{1}{4}(x+y)\left[(x+y)^2+3(x-y)^2\right]=2.\;$ It follows that

\displaystyle\begin{align} x^2+y^2&=\frac{1}{2}\left[(x+y)^2+(x-y)^2\right]\\ &=\frac{1}{3}(x+y)^2+\frac{4}{3}\frac{1}{x+y}\\ &=\frac{1}{3}\left(1+\frac{1}{x+y}\right)(x+y-2)^2+(x+y)\\ &\ge x+y. \end{align}

### Illustration ### Modification 1

Let $x,y\gt 0.\;$ Assume $x+y=2.\;$ Prove that

$x^3+y^3\ge x^2+y^2.$

From $(x+y)^3=x^3+y^3+3xy(x+y)\;$ we get $x^3+y^3=8-6xy.\;$ From $(x+y)^2=x^2+y^2+2xy,\;$ $x^2+y^2=4-2xy.\;$ Thus $x^3+y^3\ge x^2+y^2\;$ is equivalent to $8-6xy\ge 4-2xy,\;$ or, $1\ge xy.$ Since by the AM-GM inequality $\sqrt{xy}\le\displaystyle\frac{x+y}{2}=1,\;$ $1\ge xy\;$ is indeed true.

### Modification 2

Let $x,y\gt 0.\;$ Assume $x^2+y^2=2.\;$ Prove that

$x^3+y^3\ge x+y.$

Multiplying $x^2+y^2=2\;$ by $x+y\;$ we get $x^3+y^3+xy(x+y)=2(x+y),\;$ or $x^3+y^3=(x+y)(2-xy)\ge x+y,\;$ for, $2-xy\ge 1,\;$ because $\displaystyle xy\le\frac{x^2+y^2}{2}=1.$

### Generalization

Let $x,y\gt 0.\;$ Assume $m,n,p\gt 1\;$ are real numbers, with $m\gt n.\;$ Finally, assume $x^p+y^p=2.\;$ Prove that

$x^m+y^m\ge x^n+y^n.$

Consider the function $f(u)=x^u+y^u,\;$ with fixed $x\;$ and $y.\;$ If $x=y=1,\;$ there is nothing to prove. Otherwise, since, for $u\ge 1,\;$ $\displaystyle\sqrt[\displaystyle u]{\frac{x^u+y^u}{2}}\ge\frac{x+y}{2},\;$ we always have $x+y\le 2,\;$ with one of $x,y$ less, the other greater, than $1.\;$ Function $f(u)\;$ is strictly increasing for $u\gt 1.$ Copyright © 1996-2018 Alexander Bogomolny

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