I would like to solve the following problem.

Let $a,b \in \mathbb{R}^+$ such that $a+2b=1$. Find the minimum value of $\frac{1}{a}+\frac{1}{b}$.

Using AM-GM, I get $$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right) \geq \sqrt{\frac{1}{ab}}$$

I tried using the constraint to rewrite as $$\frac{1}{2}\left(\frac{1}{1-2b}+\frac{1}{b}\right) \geq \sqrt{\frac{1}{(1-2b)b}}$$

However, I don't see how this would help me as the right side of the inequality does not reduce to a constant.

I have seen AM-GM examples utilizing creative tricks to create a constant and I wonder if there is one for this problem. Thanks!


Using Cauchy-Schwarz inequality, $$(a+2b)\left(\frac{1}{a}+\frac{1}{b}\right) \ge (1+\sqrt2)^2 \implies \frac{1}{a}+\frac{1}{b} \ge (1+\sqrt2)^2.$$ The equality holds when $$a = \sqrt{2}b \implies a=\frac{1}{1+\sqrt{2}}, b=\frac{1}{2+\sqrt{2}}$$

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    $\begingroup$ So you have a lower bound. Unless you show it can be actually achieved for allowable $a,b$, you cannot claim it’s a minimum. In this case you need to show $a+2b=1, a^2=2b^2$ has a non negative solution. $\endgroup$ – Macavity Nov 3 '17 at 4:14
  • $\begingroup$ @Macavity The lower bound is achieved when $$a=\frac{1}{1+\sqrt{2}}, b=\frac{1}{2+\sqrt{2}}.$$ $\endgroup$ – Math Lover Nov 3 '17 at 14:48

If the application of trigonometry is allowed,

WLOG. $a=\sin^2t,2b=\cos^2t$

Now use $$\dfrac{x\tan^2t+y\cot^2t}2\ge\sqrt{x\tan^2t\cdot y\cot^2t}=?$$ for $x,y>0$


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