It is minimum when a1=a2=a3=a4=a5=a6=a7=6/7
Which gives
(6/7)^7 / { 1-6/7}^7 = (6/7)^7 / (1/7)^7 =6^7
So,
ab=6*7=42
–
Saurav Shakya
·
3 years, 8 months ago

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@Saurav Shakya
–
GM-HM inequality states that for positive reals \(a_i\) then \(\sqrt[n]{a_1a_2 \ldots a_n}\geq \frac{n}{\frac {1}{a_1}+\frac{1}{a_2} \ldots \frac{1}{a_n}}\) where equality occurs when all \(a_i\) are equal. So the value is always greater or equal than \(\frac{7}{\frac {1-a_1}{a_1}+\frac{1-a_2}{a_2} \ldots \frac{1-a_n}{a_n}}\) to the power of seven and its minimum is when this equality occurs. Here I assume them as positive reals less than 1, because if not, then let some \(a_k>1\), then the value is negative and grows smaller to negative infinity as the choice of \(a_k\) grow bigger.
–
Yong See Foo
·
3 years, 8 months ago

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TopNewestIt is minimum when a1=a2=a3=a4=a5=a6=a7=6/7 Which gives (6/7)^7 / { 1-6/7}^7 = (6/7)^7 / (1/7)^7 =6^7 So, ab=6*7=42 – Saurav Shakya · 3 years, 8 months ago

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– Yong See Foo · 3 years, 8 months ago

GM-HM inequality states that for positive reals \(a_i\) then \(\sqrt[n]{a_1a_2 \ldots a_n}\geq \frac{n}{\frac {1}{a_1}+\frac{1}{a_2} \ldots \frac{1}{a_n}}\) where equality occurs when all \(a_i\) are equal. So the value is always greater or equal than \(\frac{7}{\frac {1-a_1}{a_1}+\frac{1-a_2}{a_2} \ldots \frac{1-a_n}{a_n}}\) to the power of seven and its minimum is when this equality occurs. Here I assume them as positive reals less than 1, because if not, then let some \(a_k>1\), then the value is negative and grows smaller to negative infinity as the choice of \(a_k\) grow bigger.Log in to reply

– Deep Chanda · 3 years, 8 months ago

nicely done Yong See F.Log in to reply

– Saurav Shakya · 3 years, 8 months ago

I assumed 0 < ai < 1 because if any one ai=0 and any one ai>=1 then we can get - infinityLog in to reply