Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$

Question

Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$

Here's what I did to solve this exercise:

$$\frac{a+b}2 \ge \sqrt{ab} \implies \frac12 \ge \sqrt{ab} \implies \frac14\ge ab \implies ab\le\frac14$$

$$\left(1+\frac1a\right)\left(1+\frac1b\right) = 1+\frac1b+\frac1a+\frac1{ab} \ge 9 \Rightarrow \frac{a+b+1}{ab}\ge8 \Rightarrow \frac2{ab}\ge8 \Rightarrow 2\le8ab \Rightarrow ab\le\frac14$$

I came to a known point, but I'm not sure that this is the right form. If there is any other more clear proof please show it to me.

The first identity doesn't seem right. Can you check it? Or, is it an assumption? — Michael Burr, Mar 11 '15 at 15:56
The post is still confusing. Can you edit please? Are you given $a^2+b^2=(a+b)/2$ and asked to prove the 2 inequalities in the title? — Kim Jong Un, Mar 11 '15 at 16:01
From where you stopped, $$\frac{a+b+1}{ab}+1 = \frac{2}{ab}+1 \ge \frac2{1/4}+1 = 9$$ — Macavity, Mar 11 '15 at 16:18
@seda That's better, though it is not equality, the LHS can be larger. — Macavity, Mar 11 '15 at 16:19

score 4 · Answer 1 · answered Mar 11 '15 at 15:56

4

Use Cauchy-Schwarz inequality $$\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge\left(1+\dfrac{1}{\sqrt{ab}}\right)^2\ge (1+2)^2=9$$

answered Mar 11 '15 at 15:56

math110

93,304

We didn't prove it in class. Can you find a way using the identities I showed? – seda Mar 11 '15 at 16:09

Mark Viola · Accepted Answer · 2015-03-11T17:10:11.283

1

If the inequality$$\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2$$is used, then we can deduce that $a^2+b^2\ge2ab$ must hold. This can be seen from squaring the terms on the LHS and simplifying the resulting expression.

Now, $$ab\le\frac12(a^2+b^2)=\frac{(a+b)^2-2ab}{2}=\frac12-ab$$This implies that $$2ab\le\frac12$$which in turn gives the first inequality$$ab\le\frac14$$

For the second inequality, we have $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=1+\frac{a+b+1}{ab}=1+\frac{2}{ab}$$But we just showed that $ab\le\frac14$ which means that $\frac{1}{ab}\ge 4$. Using this, we find $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge1+(2)(4)=9$$and that completes the proof for the second inequality!

edited Mar 11 '15 at 17:10

answered Mar 11 '15 at 16:28

Mark Viola

179,405

who said that a=b? We haven't learnt this method – seda Mar 11 '15 at 16:34
@seda The first condition you wrote was $\frac{a^2+b^2}{2}=\left(\frac{a+b}{2}\right)^2$. If you square the terms on the RHS and simplify, you will get $a^2+b^2-2ab=0$. But $a^2+b^2-2ab=(a-b)^2$=0. So, $a$ must be equal to $b$. – Mark Viola Mar 11 '15 at 16:42
Please check it again. I made a mistake and I corrected it. – seda Mar 11 '15 at 16:44
Of course! Thanks a lot for your work! I really appreciate it! I am given the chance just to accept answers and that's what I did. – seda Mar 11 '15 at 17:18
Thank you!!!! Let me know if you need help with other problems as they arise. – Mark Viola Mar 11 '15 at 17:20

score 1 · Answer 3 · answered Mar 11 '15 at 17:41

Subtract the inequality from the equation $$ 1=(a+b)^2=a^2+2ab+b^2\\ 0\le(a-b)^2=a^2-2ab+b^2 $$ to get $$ 1\ge4ab\implies\frac14\ge ab\tag{1} $$ Apply $(1)$ to get $$ \begin{align} \left(1+\frac1a\right)\left(1+\frac1b\right) &=1+\frac1a+\frac1b+\frac1{ab}\\ &=1+\frac{a+b}{ab}+\frac1{ab}\\[3pt] &=1+\frac2{ab}\\[3pt] &\ge1+\frac2{1/4}\\[8pt] &=9\tag{2} \end{align} $$

score 0 · Answer 4 · answered Mar 11 '15 at 16:15

If you're only allowed to use $x+y\geq 2\sqrt{xy}$ (for $x,y\geq 0$), then $$ 1=a+b\geq2\sqrt{ab}\implies ab\leq\frac{1}{4} $$ and $$ (1+1/a)(1+1/b)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\geq 1+\frac{2}{\sqrt{ab}}+\frac{1}{ab}\geq1+\frac{2}{\sqrt{1/4}}+\frac{1}{1/4}=9. $$

Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$

4 Answers4