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,

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 $a+b$?

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!!!

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## Comments

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TopNewestMy 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

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plss

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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 $f(x)=\left(x+\frac 1x\right)^2$ is convex for all $x\in\Bbb R$, so using Jensen's inequality on $f$ with $a,b$, we have,

$\sum_{x\in\{a,b\}}\left(x+\frac 1x\right)^2=f(a)+f(b)\geq 2f\left(\frac{a+b}2\right)=2f(1/2)=2\left(\frac 12+2\right)^2=2\times\frac{25}4=\frac{25}2$

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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).

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

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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!!!

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THANKS A LOT!!!

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