An Inequality with Just Two Variable VII

1. $\displaystyle x^2-xy+y^2\ge\frac{1}{2}(x^2+y^2)\ge\frac{1}{4}(x+y)^2.\,$ Also $x^2+y^2\ge 2xy.$

2. For $x,y\ge 0,\,$ $x+y\ge 2\sqrt{xy}.$

The given inequality is equivalent to

$\displaystyle (x+y)^3(x^2-xy+y^2)^3(x^2-xy+y^2)\ge x^2y^2\sqrt{xy}(x^2+y^2)^3.$

Thus,

$\displaystyle\begin{align}(x+y)^3(x^2-xy+y^2)&\ge\frac{1}{4}(x+y)^4(x+y)\\ &\ge 8x^2y^2\sqrt{xy},\\ (x^2-xy+y^2)^3&\ge\frac{1}{8}(x^2+y^2)^3 \end{align}$

Multiplying the two gives the require inequality.

By Chebyshev's inequality,

$\displaystyle x^3+y^3\ge\frac{1}{2}(x+y)(x^2+y^2).$

By the AM-GM inequality,

$\displaystyle x^2-xy+y^2\ge xy.$

Thus

$\displaystyle (x^3+y^3)^3(x^2-xy+y^2)\ge\frac{xy}{8}(x+y)^2(x^2+y^2)^3\ge RHS$

Because,

$\displaystyle (x+y)^3\ge (2\sqrt{xy})^3=8xy\sqrt{xy}.$

Multiply the require inequality by $(x+y)\,$ which reduce it to:

$\displaystyle (x^3+y^3)^4\ge x^2y^2\sqrt{xy}(x^2+y^2)^3(x+y).$

We have

$\displaystyle \begin{align} (x^3+y^3)^4 &\ge \left(\frac{1}{2}(x+y)(x^2+y^2)\right)^4\\ &=\left(\frac{x+y}{2}\right)^2\cdot \left(\frac{x^2+y^2}{2}\right)\cdot\frac{(x+y)^2}{2}\cdot (x^2+y^2)^3\\ &\ge (xy)^2\cdot \frac{x+y}{2}\cdot (x+y)\cdot (x^2+y^2)^3\\ &\ge (xy)^2\sqrt{xy}(x^2+y^2)^3(x+y). \end{align}$

If $x=0\,$ or $y=0,\,$ there is nothing to prove. Assume $x,y\gt 0.\,$ Set $x=r\cos\theta,\,$ $y=r\sin\theta.\,$ The required inequality reduces to

$r^{11}(\cos^3\theta+\sin^3\theta)^3(\cos^2\theta-\cos\theta\sin\theta+\sin^2\theta)\ge r^{11}(\cos\theta\sin\theta)^{\frac{5}{2}},$

or,

$(\cos\theta+\sin\theta)^3(1-\cos\theta\sin\theta)^4\ge (\cos\theta\sin\theta)^{\frac{5}{2}}.$

By the AM-GM inequality, $\cos\theta+\sin\theta\ge 2\sqrt{\cos\theta\sin\theta}.$ Also, since $2\cos\theta\sin\theta=\sin 2\theta\le 1,\,$ $1-\cos\theta\sin\theta\ge\cos\theta\sin\theta.\,$ In addition,

$\displaystyle \begin{align} 1-\cos\theta\sin\theta &= 1-\frac{1}{2}\left[1-(\cos\theta-\sin\theta)^2\right]\\ &=\frac{1}{2}\left[1+(\cos\theta-\sin\theta)^2\right] \end{align}$

It follows that

$\displaystyle \begin{align} LHS&\ge 8(\cos\theta\sin\theta)^{\frac{3}{2}}(\cos\theta\sin\theta)\times\frac{1}{8}\left[1+(\cos\theta-\sin\theta)^2\right]^3\\ &=(\cos\theta\sin\theta)^{\frac{5}{2}}[\left[1+(\cos\theta-\sin\theta)^2\right]^3\\ &\ge (\cos\theta\sin\theta)^{\frac{5}{2}}=RHS. \end{align}$

From $\displaystyle \sqrt[3]{\frac{x^3+y^3}{2}}\ge\sqrt{\frac{x^2+y^2}{2}},\,$ we have

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

Also, $x^3+y^3\ge 2\sqrt{x^3y^3}\,$ and $x^2-xy+y^2\ge xy.\,$ Multiplying all three gives the required inequality.

The require inequality is equivalent to

$(x^3+x^3)^4\ge (xy)^{\frac{5}{2}}(x^2+y^2)^3(x+y).$

Now,

$\displaystyle \begin{align} 2(x^3+y^3)^2&\ge (x^2+y^2)^3\\ x^3+y^3&\ge 2(xy)^{\frac{3}{2}}\\ x^3+y^3&\ge \frac{1}{2}(x^2+y^2)(x+y)\\ &\ge\frac{1}{2}\cdot 2\sqrt{x^2y^2}(x+y)\\ &=xy(x+y). \end{align}$

The product of the three yields the required inequality.

This problem from his book "Algebraic Phenomenon" has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Seyran Ibrahimov; Solution 3 by Myagmarsuren Yadamsuren; Solution 4 by Ravi Prakash.