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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
3 months 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). · 3 months ago

thanks for yout help but i think it should (a^2+b^2)+(1/a^2+1/b^2) · 3 months ago

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!!! · 3 months ago

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,

$\sum_{x\in\{a,b\}}\left(x+\frac 1x\right)^2=f(a)+f(b)\geq 2f\left(\frac{a+b}2\right)=2f(1/2)=2\left(\frac 12+2\right)^2=2\times\frac{25}4=\frac{25}2$ · 3 months ago

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. · 3 months ago

can u use cauchy-schwarz inequality because this question was after this theorem · 3 months ago

QM-AM is a type of Cauchy-schwarz inequality. · 3 months ago

plss · 3 months ago