3

let $a,b,c>0$ and such $abc=1$,show that $$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$

my idea: maybe can use AM-GM inequality, $$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$ $$\Longleftrightarrow 3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$$

then I can't,Thank you very much

math110
  • 93,304

2 Answers2

3

Here is a straightforward (but not very elegant) solution : let us put

$$ D=3(a^4+b^4+c^4)+2(a+b+c)abc - 5(a^2b^2+b^2c^2+a^2c^2) $$

We may assume without loss that $a\leq b\leq c$. Then, if we put $u=b-a$ and $v=c-b$, we have

$$ D=u^4 + (8a + 2v)u^3 + (10a^2 + 12va + 13v^2)u^2 + (10va^2 + 28v^2a + 12v^3)u + (10v^2a^2 + 12v^3a + 3v^4) $$

and we are done since all the coefficients are positive.

Ewan Delanoy
  • 61,600
3

Consider Schur's inequality for $r = 2$: $$a^4 + b^4 + c^4 + (a + b + c)abc \geqslant a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$

plus three AM-GMs: $$ \begin{aligned} a^3b + b^3a &\geqslant 2a^2b^2\\ b^3c + c^3b &\geqslant 2b^2c^2\\ a^3c + c^3a &\geqslant 2a^2c^2 \end{aligned} $$

This gives you, after multiplying by $2$: $$2(a^4 + b^4 + c^4) + 2(a + b + c)abc \geqslant 4(a^2b^2 + b^2c^2 + a^2c^2).$$

Three more AM-GMs: $$ \begin{align} \frac{a^4 + b^4}{2} &\geqslant a^2b^2,\\ \frac{b^4 + c^4}{2} &\geqslant b^2c^2,\\ \frac{a^4 + c^4}{2} &\geqslant a^2c^2. \end{align}$$

Sum up 4 last inequalities and you're done.

Equality, of course, holds only if $a = b = c$, which can be seen from all those AM-GMs.

ElThor
  • 912