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?
Thanks for your help!

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