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!

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!
