$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $

Question

$x,y,z > 0$, prove that $$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$

Without using Schur's inequality,

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Attempt:

By $C.S$:

$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} = x^{3/2}y^{3/2} + y^{3/2}z^{3/2} + z^{3/2}x^{3/2} $$ $$ \le \sqrt{ x^{3} + y^{3} +z^{3}} \sqrt{ y^{3} + z^{3} + x^{3}} = x^{3} + y^{3} + z^{3} $$

then I have to prove $$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \le 3xyz$$

this one is difficult. Another thing that we know by AM-GM: $ x^{3} + y^{3}+ z^{3} \ge 3xyz$.

Answer 1

First, Schur's inequality provides that $$ x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x). $$ Then, $$ xy(x+y)\ge 2xy\sqrt{xy} $$ and hence $$ x^3+y^3+z^3+3xyz\ge 2\big(xy\sqrt{xy} +yz\sqrt{yz} +zx\sqrt{zx} \big) $$

Answer 2

Let $x=e^a$, $y=e^b$, $z=e^c$ $\;$

and $a\ge b \ge c$ (WLOG)

Our inequality is equivalent to $$3e^{a+b+c} + e^{3a}+ e^{3b} + e^{3c} \ge 2 \left( e^{3(a+c)/2} + e^{3(a+c)/2} + e^{3(a+b)/2} \right)$$

If $a+c\le 2b$ ,

This inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec \left( 3a,3b,(a+b+c),(a+b+c),(a+b+c),3c \right)$$

The majorization holds because,

$$3a\ge \frac{3(a+b)}{2}$$

$$3a+ 3b \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$

$$3a+ 3b + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$$3a+ 3b + (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$$3a+ 3b + (a+b+c) + (a+b+c)+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}+ \frac{3(b+c)}{2}$$

$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$

$3^{rd}$ and $4^{th}$ inequalities by $a+ c \le 2b$. $\;$ Other inequalities are obvious.

If $a+c\ge 2b$ ,

The desired inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec \left( 3a,(a+b+c),(a+b+c),(a+b+c),3b,3c \right)$$

The majorization holds because,

$$3a\ge \frac{3(a+b)}{2}$$

$$3a+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$

$$3a+ (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$

$\dots$

$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$

$3^{rd}$ inequality comes from $a+ c \ge 2b$. $\;$ Other inequalities are obvious.