MHB Maximum Value of k in $(1+\frac{1}{a})(1+\frac{1}{b}) \ge k$ is 9

[Mathick]

For any positive real numbers a, b such that $a+b=1$, the inequality $(1+\frac{1}{a})(1+\frac{1}{b})\ge k$. Find the maximum value of $k \in \Bbb{R}$.

I did:

$(1+\frac{1}{a})(1+\frac{1}{b})= \frac{a+1}{a} \cdot \frac{b+1}{b}=\frac{ab+a+b+1}{ab}= 1+\frac{2}{ab} \ge k$

Using the Cauchy's inequality, we have

$\frac{a+b}{2} \ge \sqrt{ab} \implies \frac{1}{4} \ge ab \implies \frac{1}{ab} \ge 4 \implies \frac{2}{ab} \ge 8$

Thus,

$(1+\frac{1}{a})(1+\frac{1}{b})= 1+\frac{2}{ab} \ge 1+ 8 = 9$

Therefore maximum value of k is 9. Is it correct?

 

[Euge]

Hi Mathick,

The answer is correct, but you should give values of $a$ and $b$ that make the equality hold. Letting $a = b = 1/2$ will do.

 

[earboth]

For any positive real numbers a, b such that $a+b=1$, the inequality $(1+\frac{1}{a})(1+\frac{1}{b})\ge k$. Find the maximum value of $k \in \Bbb{R}$.

...

And I did:

Substitute b = 1-a and you'll get

$$k(a)\leq \frac{(a+1)(a-2)}{a(a-1)}$$

Calculate the extremum by using the 1^st derivation:

$$\frac{dk}{da}\leq \frac{2(2a-1)}{a^2(a^2-1)}$$

which is zero if $$a = \frac12$$. Determine b.

 

[Mathick]

And I did:

Substitute b = 1-a and you'll get

$$k(a)\leq \frac{(a+1)(a-2)}{a(a-1)}$$

Calculate the extremum by using the 1^st derivation:

$$\frac{dk}{da}\leq \frac{2(2a-1)}{a^2(a^2-1)}$$

which is zero if $$a = \frac12$$. Determine b.

Well... but how to find a value of k now?

 

[earboth]

Well... but how to find a value of k now?

Good morning,

replace a by $$\frac12$$ in

$$k(a)\leq \frac{(a+1)(a-2)}{a(a-1)}$$

and you'll get

$$k\left(\frac12 \right) \leq \frac{\left(\frac12 +1 \right)\left(\frac12 - 2 \right)}{\frac12 \left(\frac12 - 1 \right)} = \frac{-\frac94}{-\frac14}=9$$

That is the maximum value for k.