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Inequalities

If \(a\) and \(b\) are positive numbers satisfying \(a+b=1\), prove that \( \left( a + \dfrac1a \right)^2 + \left( b + \dfrac 1b\right)^2 \geq \dfrac{25}2 \).

Note by Subham Subian
9 months, 2 weeks ago

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Hint: This is equivalent to proving that \((a^2 + b^2) + \left( \dfrac1a + \dfrac1b \right) \geq 8.5 \).

Now, for positive \(a\) and \(b\), what is the relationship between \(a^2 + b^2\) and \(a+b\)? Similarly, what is the relationship between \(\dfrac1a + \dfrac1b \) and \(a+b\)?

Read up Power mean inequality (qagh). Pi Han Goh · 9 months, 2 weeks ago

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@Pi Han Goh thanks for yout help but i think it should (a^2+b^2)+(1/a^2+1/b^2) Subham Subian · 9 months, 1 week ago

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@Subham Subian cause min. value of a^2+b^2 IS 1/2 and we need max.value of ab which is1/4 but after computing we get >=4.5 but if u take (a^2+b^2)+(1/a^2+1/b^2) we get the right answer ANYWAY THANKS FOR YOUR HELP!!!!!!!!!! THANKS A LOT!!! Subham Subian · 9 months, 1 week ago

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Another approach would be to note that the function \(f(x)=\left(x+\frac 1x\right)^2\) is convex for all \(x\in\Bbb R\), so using Jensen's inequality on \(f\) with \(a,b\), we have,

\[\sum_{x\in\{a,b\}}\left(x+\frac 1x\right)^2=f(a)+f(b)\geq 2f\left(\frac{a+b}2\right)=2f(1/2)=2\left(\frac 12+2\right)^2=2\times\frac{25}4=\frac{25}2\] Prasun Biswas · 9 months, 2 weeks ago

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My latex is not working properly so I will just give a brief of the solution.

Apply AM-GM to get maximum value of ab. Apply QM-AM and substitute maximum value of ab to get minimum value of expression as ab is in denominator. Svatejas Shivakumar · 9 months, 2 weeks ago

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@Svatejas Shivakumar can u use cauchy-schwarz inequality because this question was after this theorem Subham Subian · 9 months, 2 weeks ago

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@Subham Subian QM-AM is a type of Cauchy-schwarz inequality. Svatejas Shivakumar · 9 months, 2 weeks ago

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@Subham Subian plss Subham Subian · 9 months, 2 weeks ago

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THANKS A LOT!!! Subham Subian · 9 months, 2 weeks ago

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