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Mathick
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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?
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?
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