Let $x,y,z$ be three positive numbers. Can you prove the follwing inequality : $({x}^{2}{y}^{2}+{z}^{4}{)}^{3}\le ({x}^{3}+{y}^{3}+{z}^{3}{)}^{4}$

Mauricio Mathis
2022-07-17
Answered

Let $x,y,z$ be three positive numbers. Can you prove the follwing inequality : $({x}^{2}{y}^{2}+{z}^{4}{)}^{3}\le ({x}^{3}+{y}^{3}+{z}^{3}{)}^{4}$

You can still ask an expert for help

eri1ti0m

Answered 2022-07-18
Author has **11** answers

We need to show $({x}^{2}{y}^{2}+{z}^{4}{)}^{3}\le ({x}^{3}+{y}^{3}+{z}^{3}{)}^{4}.$

By AM-GM, we have $({x}^{3}+{y}^{3}+{z}^{3}{)}^{4}\ge {(2(xy{)}^{\frac{3}{2}}+{z}^{3})}^{4}$

Let $a=\sqrt{xy}>0$. Then it is sufficient to show that$(2{a}^{3}+{z}^{3}{)}^{4}\ge ({a}^{4}+{z}^{4}{)}^{3}$

Let $t=\frac{a}{z}>0$, then we need to show

$f(t)=(2{t}^{3}+1{)}^{4}-({t}^{4}+1{)}^{3}\ge 0$ for $t>0$

or $f(t)={t}^{3}(15{t}^{9}+32{t}^{6}-3{t}^{5}+24{t}^{3}-3t+8)>0$

Now, note that for $t\ge 1$

$3{t}^{6}-3{t}^{5}=3{t}^{5}(t-1)$ and $3{t}^{3}-3t=3t({t}^{2}-1)$

so $f(t)={t}^{3}[15{t}^{9}+32{t}^{6}+3{t}^{3}(1-{t}^{2})+21{t}^{3}+3(1-t)+5]>0.$

Thus $f(t)>0$ when t>0, and hence the inequality holds.

By AM-GM, we have $({x}^{3}+{y}^{3}+{z}^{3}{)}^{4}\ge {(2(xy{)}^{\frac{3}{2}}+{z}^{3})}^{4}$

Let $a=\sqrt{xy}>0$. Then it is sufficient to show that$(2{a}^{3}+{z}^{3}{)}^{4}\ge ({a}^{4}+{z}^{4}{)}^{3}$

Let $t=\frac{a}{z}>0$, then we need to show

$f(t)=(2{t}^{3}+1{)}^{4}-({t}^{4}+1{)}^{3}\ge 0$ for $t>0$

or $f(t)={t}^{3}(15{t}^{9}+32{t}^{6}-3{t}^{5}+24{t}^{3}-3t+8)>0$

Now, note that for $t\ge 1$

$3{t}^{6}-3{t}^{5}=3{t}^{5}(t-1)$ and $3{t}^{3}-3t=3t({t}^{2}-1)$

so $f(t)={t}^{3}[15{t}^{9}+32{t}^{6}+3{t}^{3}(1-{t}^{2})+21{t}^{3}+3(1-t)+5]>0.$

Thus $f(t)>0$ when t>0, and hence the inequality holds.

asked 2020-10-28

Find a polynomial with integer coefficients that satisfies the given conditions. Q has degree 3 and zeros 0 and i.

asked 2021-02-25

Find a polynomial of the specified degree that has the given zeros. Degree 4, zeros -2, 0, 2, 4

asked 2022-10-31

Consider the following polynomial of x

${x}^{2}+(a+b\phantom{\rule{thinmathspace}{0ex}}g)x+c\phantom{\rule{thinmathspace}{0ex}}{g}^{2}=0$

where $b\in \mathbb{R}$, and $c\in \mathbb{R}$, whereas g can be complex. $c\ne 0$

${x}^{2}+(a+b\phantom{\rule{thinmathspace}{0ex}}g)x+c\phantom{\rule{thinmathspace}{0ex}}{g}^{2}=0$

where $b\in \mathbb{R}$, and $c\in \mathbb{R}$, whereas g can be complex. $c\ne 0$

asked 2022-05-07

$(6{s}^{2}+6s+7)\ufeff+(7s-3{s}^{2})=$?

asked 2022-07-24

True or False?

$2{x}^{3}\u20133{x}^{4}+4{x}^{5}$ is a polynomial function of degree 3.

$2{x}^{3}\u20133{x}^{4}+4{x}^{5}$ is a polynomial function of degree 3.

asked 2022-08-15

How to find the roots of ${x}^{4}+1$

asked 2022-09-20

If I have a polynom p with $p(a)=0$, how to construct a polynom q with $q({a}^{-1})=0$