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Let $a,b\ge0: a^4+b^4=17$. Prove that: $$15(a+b)\ge17+14\sqrt{2ab}$$

I am looking for a nice approach. My approach is ugly by replace: $b=\sqrt[4]{17-a^4} $ and the rest is working with fuction.

I am quite sure $(1,2)$ is the only case equality so I guess we can use AM-GM in someway. Is there any better idea for this problem?

Thanks for your help!

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    $\begingroup$ An interesting note: If you replace the condition by $a^r+b^r\leq 2^r+1$ for any $1\leq r<4$, or if you decrease the numbers $15$, $17$, and $14$ by the same fixed amount $\epsilon$, the inequality seems to fail. $\endgroup$ Dec 22, 2021 at 5:58
  • $\begingroup$ Very nice. I will notice that $\endgroup$
    – Mars
    Dec 22, 2021 at 12:42

3 Answers 3

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Marginally less ugly, let $s=a+b, p=ab$, then:

$$ a^4+b^4 = 17 \;\;\iff\;\; (s^2-2p)^2-2p^2=17 \;\;\iff s^4 - 4ps^2 + 2p^2 - 17 = 0 $$

Solving the quadratic in $\,p\,$, and retaining the root which satisfies $p \le s^2$:

$$ p = s^2 - \sqrt{\frac{s^4+17}{2}} $$

The inequality to prove is equivalent to:

$$ (15s-17)^2 \ge 2\cdot 14^2 \,p = 14^2\left(2s^2-\sqrt{2\left(s^4+17\right)}\right) $$

Rearranging with positive quantities on both sides and squaring:

$$ 2 \cdot 14^4 \left(s^4+17\right) \ge \left(2 \cdot 14^2 s^2 - (15s-17)^2\right)^2 $$

After expanding, collecting and "luckily" finding the rational root $s=3$:

$$ 17 (2879 s^4 - 10020 s^3 - 9622 s^2 + 17340 s + 71919) \ge 0 \\ \iff\;\;\;\; (s - 3)^2 (2879 s^2 + 7254 s + 7991) \ge 0 $$

The quadratic factor has no real roots, so the inequality holds true, with equality iff $s=3\,$, which then gives $p=2$ i.e. $\{a,b\}=\{1,2\}$.

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  • $\begingroup$ Indeed, it is nicer my approach. Thanks! $\endgroup$
    – Mars
    Dec 22, 2021 at 12:10
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Due to Erik Satie's hint:

We will prove that: $$\frac{17^2}{a^4+b^4}+60(a+b)\ge 85+56\sqrt{2ab}$$ By AM-GM: $$\frac{17}{a^4+b^4}\ge 5-4\sqrt[4]{\frac{a^4+b^4}{17}}\ge 5-\frac{4}{\sqrt{17}}\sqrt{5(a^2+b^2)-4ab}$$ The inequality becomes: $$15(a+b)\ge\sqrt{17\left(5(a^2+b^2)-4ab\right)}+14\sqrt{2ab}$$ By C-S inequality, it turns out: $$17\sqrt{\frac{5(a^2+b^2)-4ab}{17}}+28\sqrt{\frac{ab}{2}}\leq\sqrt{45\left(5(a^2+b^2)-4ab+14ab\right)}=15(a+b)$$ Obviously, equality holds iff $(a,b)=(1,2); (2,1)$. The proof is done!

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This is not an answer.

I don't know if it's a right approach but let me propose it :

First we substitute :

$$a^4=x^4,\quad b^4=4y^4$$

Then we use The Sophie Germain identity to factor :

$$x^{4}+4y^{4}-\left(\left(x+\sqrt{2}y\right)^{2}+\left(2-2\sqrt{2}\right)xy\right)\left(\left(x+\sqrt{2}y\right)^{2}+\left(-2-2\sqrt{2}\right)xy\right)=0$$

Then we use the fact that :

$$17=1+4\left(\sqrt{2}\right)^{4}=\left(5-2\sqrt{2}\right)\left(5+2\sqrt{2}\right)$$

Now I want to put :

$$\frac{\left(\left(x+\sqrt{2}y\right)^{2}+\left(2-2\sqrt{2}\right)xy\right)}{c}=\left(5+2\sqrt{2}\right)$$

And :

$$c\left(\left(x+\sqrt{2}y\right)^{2}+\left(-2-2\sqrt{2}\right)xy\right)=\left(5-2\sqrt{2}\right)$$

But I have a doubt on it so I stop here .

Hope you find something relevant .

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  • $\begingroup$ Thanks. Due to yourt hint, I get nice proof by AM-GM $\endgroup$
    – Mars
    Dec 22, 2021 at 11:28
  • $\begingroup$ @Mars Hum okay thanks for the nice problem ^_^. $\endgroup$
    – Erik Satie
    Dec 22, 2021 at 11:38
  • $\begingroup$ Btw, please help me prove this: math.stackexchange.com/questions/4335241/… $\endgroup$
    – Mars
    Dec 22, 2021 at 13:12

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