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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!

amWhy
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Sickness
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    Hint: You're trying to bound a square ($b^2$) by a linear term $\sqrt{2(1/a^2+a^2)}b$, so that suggests to consider this as a quadratic in $b$ and show that the discriminant is non-positive. (And if it was positive, then show that the larger root is still negative. Luckily that doesn't happen here.) Equality holds when the discriminant is 0, and find the corresponding root. – Calvin Lin Dec 20 '21 at 19:06
  • Very effective dear friend. Now I get it, thank you so much – Sickness Dec 21 '21 at 01:26
  • Can you write it up as an answer? – Calvin Lin Dec 21 '21 at 16:28

2 Answers2

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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: $$\frac{\frac{1}{a^2}+b^2}{\sqrt{\frac{1}{a^2}+a^2}}\ge\sqrt{2}(b-a+1)$$

Sickness
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    Wow, this appears crazy to me / hard to motivate why we should do that. What was your thinking? – Calvin Lin Dec 22 '21 at 20:25
  • Thanks, I learnt this technique from my teacher Jokehim. It is called Setting equality cases base on lower bound terms – Sickness Dec 23 '21 at 01:23
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Since OP posted their own version, I'd post my solution.

WTS $ b^2 - \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot b + [\frac{1}{a^2} + \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot (a-1) ] \geq 0$.

We will prove this by viewing it as a quadratic in $b$, and showing that the discriminant is $ \leq 0$ for $ a > 0$, hence the value is always $ \geq 0 $.

$$D = [ \sqrt{2 ( \frac{1}{a^2 } + a^2 )}] ^2 - 4[\frac{1}{a^2} + \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot (a-1) ] \\ = - \frac{ 2 ( a-1) ( a^3+a^2+a+1 - 2\sqrt{2} \sqrt{ a^4 + 1 }) }{a^2} .$$

Observe that $ ( a^3 + a^2 + a + 1 ) ^2 - 8 (a^4 + 1) = (a-1) ( a^5 + 3a^4 - 2a^3 + 2a^2 + 5a + 7) $.
Clearly the second term is positive for $ a > 0$.

Hence

  • If $ a > 1 $, then $ a^3 + a^2 + a + 1 > \sqrt{ 8 (a^4 + 1)}$, so $ D < 0 $
  • If $ a = 1$, then $ D = 0 $
  • If $ 0 < a < 1$, then $a^3 + a^2 + a + 1 < \sqrt{ 8 (a^4 + 1)},$ so $ D < 0 $.

Equality holds when $ D = 0 \Rightarrow a = 1 $ with corresponding root $b = 1$.

Modification of OP's solution.

Lemma: $$\frac{ \sqrt{2} x^2 } { \sqrt{ \frac{1}{a^2} + a^2 } } + \frac{ a^2 - a + 1 } { a } \geq 2 x . $$

Proof: We apply AM-GM directly. It remains to show that

$$ \sqrt{ \frac{ 2a^2} { a^4 +1 }} \times { \frac{ a^2 - a + 1 } { a } } \geq 1$$

or that $2 ( a^2 - a + 1 ) ^2 \geq (a^4 + 1) $, which is true since $2 ( a^2 - a + 1 ) ^2 - (a^4 + 1) = (a-1)^4$.
Equality holds when $ a = 1, x = 1 $.

Corollary: Apply the lemma to $ x = \frac{1}{a}$ and $x = b$, sum them up and shift terms around to get the desired inequality.
Equality holds when $ a = 1, b = 1 $.

Notes

  • (I thought that) My contribution here is to simplify the AM-GM that OP did.
    • Since $ \frac{\frac{1}{a} + a +\sqrt{2} } { 2 + \sqrt{2} } + \frac{ \frac{ 1}{a} + a - \sqrt{2} } { 2 - \sqrt{2} } = 2\frac{ a^2 - a + 1 } { a } $, arguably this made it more complicated.
  • It's hard to motivate the lemma, which was obtained from OP's AM-GM.
Calvin Lin
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