# prove: $15(a+b)\ge17+14\sqrt{2ab}$

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?

• 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. Dec 22, 2021 at 5:58
• Very nice. I will notice that
– Mars
Dec 22, 2021 at 12:42

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. $$\{a,b\}=\{1,2\}$$.

• Indeed, it is nicer my approach. Thanks!
– Mars
Dec 22, 2021 at 12:10

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!

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 .

• Thanks. Due to yourt hint, I get nice proof by AM-GM
– Mars
Dec 22, 2021 at 11:28
• @Mars Hum okay thanks for the nice problem ^_^. Dec 22, 2021 at 11:38