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I change $\frac{1}{a}+\frac{8}{b}=1$ into $\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2}=1$, then I get $a^2+b^2=(a^2+b^2)(\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2})=65+\frac{16a}{b}+\frac{16b}{a}+\frac{64a^2}{b^2}+\frac{b^2}{a^2}$

though I get $\frac{16a}{b}+\frac{16b}{a}\geq 32$,$\frac{64a^2}{b^2}+\frac{b^2}{a^2}\geq 16$, their equivalency condition is not identical.

How can I solve it?

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    $\begingroup$ Holders Inequality will yield least value as $125$ when $a=5,b=10$ $\endgroup$ Apr 6, 2023 at 4:25

3 Answers 3

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Solution: The minimum value is $125$ at $a = 5$ and $b = 10$.

Approach: First let's express $b$ in terms of $a$. From $\frac{1}{a} + \frac{8}{b} = 1$, we have

$$1 - \frac{1}{a} = \frac{8}{b} \implies \frac{a - 1}{8a} = \frac{1}{b}$$ $$\implies b = \frac{8a}{a-1}$$

and now since $b$ must be $> 0$, we have $\frac{8a}{a - 1} > 0 \implies a > 1$ which is stronger than $a > 0$ (from problem statement).

Thus, all that is left to find is the minimum of the equation $a^2 + b^2$ or, $f(a) = a^2 + \frac{64 a^2}{(a - 1)^2}$ where $a>1$. This minimum (which happens to be 125) can be easily found via differentiation.

Differentiation: The required minima will be obtained when $\frac{df(a)}{da} = 0$ where $a > 0$.

$$\frac{df(a)}{da} = 0,\ \text{where } a > 0$$ $$\implies \frac{2a(a-5)(a(a + 2) + 13)}{(a-1)^3} = 0,\ \text{where } a > 0$$ $$\implies (a - 5) > 0 \implies a = 5$$ $$\implies a = 5,\ b= \frac{8\times 5}{4} = 10$$

Thus, the minimum is obtained at $a = 5, b = 10$ and the value is $a^2 + b^2 = 25 + 100 = 125$ as needed.

Note that I have skipped some steps (when coming to $a - 5 > 0$) and arguments (prove that the point $a = 5$ is indeed a minima for $f(a)$ using $f''(a)$ at $a = 5$) here.

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  • $\begingroup$ Your right, nice solution $\endgroup$
    – QC_QAOA
    Apr 6, 2023 at 4:19
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    $\begingroup$ @NoChance for $a = 3$ and $a = 9$, $\frac{df(a)}{da}$ is not zero right? They are not stationary points. So you won't find a global (or local) minima at those two points. $\endgroup$
    – Arjo
    Apr 6, 2023 at 5:18
  • $\begingroup$ You are correct. Thanks. $\endgroup$
    – NoChance
    Apr 6, 2023 at 5:29
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Hint: Using Hölder's inequality $$\left(a^2+b^2 \right)\left(\frac1a+\frac8b \right)^2\geqslant (1+4)^3$$ with equality possible when $a^2:\dfrac1a=b^2:\dfrac8b$

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    $\begingroup$ wow! that's right! $\endgroup$
    – S.Y.Li
    Apr 6, 2023 at 4:50
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    $\begingroup$ How do you use Holder's inequality here? Do you take $p = q = 2\ $? What is your $x_1,x_2,y_1$ and $y_2\ $? What I know is that if $\frac {1} {p} + \frac {1} {q} = 1$ then Holder's inequality says that $\sum\limits \left \lvert x_i y_i \right \rvert \leq \left (\sum\limits |x_i|^p \right )^{\frac {1} {p}} \left (\sum\limits |y_i|^q \right )^{\frac {1} {q}}.$ $\endgroup$ Apr 6, 2023 at 5:04
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    $\begingroup$ Oh! I got it. You take $p = 3$ and $q = \frac {3} {2}$ and $x_1 = a^{\frac {2} {3}}, x_2 = b^{\frac {2} {3}}, y_1 = \frac {1} {x_1}$ and $y_2 = \frac {4} {x_2}.$ $\endgroup$ Apr 6, 2023 at 5:33
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    $\begingroup$ @AnilBagchi. This is a simpler version of the same inequality, essentially as you noted, $p=3, q = \frac32$ in the original form. $\endgroup$
    – Macavity
    Apr 6, 2023 at 5:37
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Another approach is to let $a = x+1, b=8(y+1)$ which gives $$\frac 1{x + 1} + \frac{1}{y + 1} =1$$

which is just $y = 1/x$. This can be parameterized by $(x, y) = p(t) = (e^{-t}, e^{t})$.

And the original constraints $a > 0, b > 0$ implies $a > 1, b > 8$ which is equivalent to $x > 0, y > 0$ which is implied by the parameterization.

Then it becomes minimizing $f(t) = (e^{-t} + 1)^2 + 64(e^t + 1)^2$ which is straightforward to work out, gives the same result as previously posted answers.

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