$a,b>0$ then: $\frac{1}{a^2}+b^2\ge\sqrt{2(\frac{1}{a^2}+a^2)}(b-a+1)$ 
Let $a,b>0$. Prove that: $$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$

Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:
Case 1: $b-a+1>0$ then squaring both side, we get equivalent inequality: $$\frac{1}{a^4}+b^4+2\frac{b^2}{a^2}\ge2\left(\frac{1}{a^2}+a^2\right)(b^2+a^2+1-2ab-2a+2b)$$ Or: $$\frac{1}{a^4}+b^4\ge\frac{2}{a^2}(a^2+1-2ab-2a+2b)+2a^2(b^2+a^2+1-2ab-2a+2b)$$
The rest is so complicated. Is there nice idea etc: AM-GM, C-S to prove this inequality.
Case 2: $b-a+1<0$ which is obviously true.
I hope we can find a better approach for the inequality. Thank you very much!

Due to @Calvin Lin, I will post solution based on his hint later.
Now I get solution by AM-GM. Notice that: $\frac{1}{a^2}+a^2=\left(\frac{1}{a}+a\right)^2-2=\left(\frac{1}{a}+a+\sqrt{2}\right)\left(\frac{1}{a}+a-\sqrt{2}\right)$
Then using AM-GM: $$2\frac{\sqrt{2}\frac{1}{a^2}}{\sqrt{\frac{1}{a^2}+a^2}}+\frac{\frac{1}{a}+a+\sqrt{2}}{2+\sqrt{2}}+\frac{\frac{1}{a}+a-\sqrt{2}}{2-\sqrt{2}}\ge\frac{4}{a}$$ Or:$$\frac{\sqrt{2}\frac{1}{a^2}}{\sqrt{\frac{1}{a^2}+a^2}}+a\ge\frac{1}{a}+1$$
Similarly, $$\frac{\sqrt{2}b^2}{\sqrt{\frac{1}{a^2}+a^2}}+a+\frac{1}{a}\ge2b+1$$
The rest is obvious: $$\frac{\frac{1}{a^2}+b^2}{\sqrt{\frac{1}{a^2}+a^2}}\ge\sqrt{2}(b-a+1)$$
