Inequalities

If aa and bb are positive numbers satisfying a+b=1a+b=1, prove that (a+1a)2+(b+1b)2252 \left( a + \dfrac1a \right)^2 + \left( b + \dfrac 1b\right)^2 \geq \dfrac{25}2 .

Note by Subham Subian
3 years 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.

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can u use cauchy-schwarz inequality because this question was after this theorem

Subham Subian - 3 years ago

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plss

Subham Subian - 3 years ago

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QM-AM is a type of Cauchy-schwarz inequality.

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

x{a,b}(x+1x)2=f(a)+f(b)2f(a+b2)=2f(1/2)=2(12+2)2=2×254=252\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 - 3 years ago

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Hint: This is equivalent to proving that (a2+b2)+(1a+1b)8.5(a^2 + b^2) + \left( \dfrac1a + \dfrac1b \right) \geq 8.5 .

Now, for positive aa and bb, what is the relationship between a2+b2a^2 + b^2 and a+ba+b? Similarly, what is the relationship between 1a+1b\dfrac1a + \dfrac1b and a+ba+b?

Read up Power mean inequality (qagh).

Pi Han Goh - 3 years ago

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thanks for yout help but i think it should (a^2+b^2)+(1/a^2+1/b^2)

Subham Subian - 3 years ago

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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 - 3 years ago

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THANKS A LOT!!!

Subham Subian - 3 years ago

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