# 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
4 years, 9 months 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.

- 4 years, 9 months ago

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

- 4 years, 9 months ago

plss

- 4 years, 9 months ago

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

- 4 years, 9 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,

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

- 4 years, 9 months ago

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).

- 4 years, 9 months ago

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

- 4 years, 9 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!!!

- 4 years, 9 months ago

THANKS A LOT!!!

- 4 years, 9 months ago