Looks so pretty, but it is too pretty

As a discussion, first consider why the "obvious" answers are wrong.
Yes, the choices were specially selected.

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Wrong Approach 1:

Applying AM-GM, we get that

$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+b} \geq 3 \sqrt[3]{ \frac{1}{ab(a+b)} } = 3 \sqrt[3]{ \frac{1}{2000}} = \frac{ 3 \sqrt[3]{4} } { 20 }.$

What is wrong with this approach?

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Wrong Approach 2:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+b} = \frac{ a+b} { ab} + \frac{1}{a+b} = \frac{ (a+b)^2} { 2000} + \frac{1}{a+b}.$

Letting $x = a+b$ , we want to minimize $\frac{ x^2}{2000 } + \frac{1}{x}$ .
Using Calculus, or AM-GM, the minimum occurs when $x = 10$ and has a value of $\frac{3}{20}$ .

What is wrong with this approach?

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Correct solution:

Let $a + b = x$ , and then $ab = \frac{2000}{x}$ . $a$ and $b$ are the real roots of the quadratic equation

$X^2 - xX + \frac{2000}{x} = 0$

This has a non-negative discriminant, so $x^2 - \frac{4 \times 2000} { x} \geq 0$ . This gives us $x^3 \geq 8000$ , or that $x \geq 20$ .

Now, we want to minimize $f(x) = \frac{ x ^2 } { 2000} + \frac{1}{x}$ SUBJECT to $x \geq 20$ . Using calculus, it is clear that for $x \geq 20$ , we have $f'(x) > 0$ . Hence, the minimum occurs when $x = 20$ .

In this scenario, we have $a + b = 20, ab = 100$ . This is satisfied when $a = 10, b = 10$ . Hence, the minimum can be achieved. This minimum is $\frac{1}{10} + \frac{1}{10} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4}$ .

By applying AM-GM $2000 = ab(a + b) \le \left( \frac{a + b}{2} \right)^2 (a + b),$ which implies $a + b \ge 20$ .

Again by the AM-GM inequality, $\begin{aligned} \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b} &= \frac{1}{2a} + \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2b} + \frac{1}{a + b} \\ &\ge 5 \sqrt[5]{\frac{1}{(2a)^2} \cdot \frac{1}{(2b)^2} \cdot \frac{1}{a + b}} \\ &= 5 \sqrt[5]{\frac{1}{16a^2 b^2 (a + b)}} \\ &= 5 \sqrt[5]{\frac{a + b}{16 a^2 b^2 (a + b)^2}} \\ &\ge 5 \sqrt[5]{\frac{20}{16 \cdot 2000^2}} \\ &= \frac{1}{4}. \end{aligned}$

Equality occurs when $a = b = 10$ , so the minimum value is 1/4.

$\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}=\frac{(a+b)^2+ab}{ab(a+b)}$

Plugging in the first equation, $\Rightarrow \frac{(a+b)^2+ab}{2000}$

Now, by A.M. G.M. inequality, the minimum value of $(a+b)^2=4ab$ , when $a=b$

Now, $\frac{5ab}{2000}=\frac{ab}{400}=\frac{a^2}{400}$

In the first equation, plugging in $a=b$

$2a^3=2000\rightarrow a^3=1000 \rightarrow a=b=10$

Therefore, $\frac{a^2}{400}=\frac{100}{400}=\boxed{\frac{1}{4}}$

I approached this problem using Lagrange Multipliers. I essentially defined the following cost function, $J$ , to be minimized with respect to the variables $a$ , $b$ , and the Lagrange Multiplier $\lambda$ :

$F_{2} (a,b) = ab(a+b) - \alpha = 0$

$F_{1} (a,b) = \frac{1}{a} + \frac{1}{b} + \frac{1}{a+b} = (a^{2} + 2ab + b^{2})(\frac{1}{\alpha})$

$J = F_{1}(a,b) + \lambda F_{2}(a,b)$ , where $\lambda$ is unknown

Taking the partial derivatives of $J$ with respect to a, b, and $\lambda$ (since we want to minimize with respect to all variables) leads to the following equations:

(1) $(2a+3b) = \alpha \lambda (2ab + b^{2})$

(2) $(2b + 3a) = \alpha \lambda (2ab + a^{2} )$

(3) $F_{2} = 0$

Now, solving for $\alpha \lambda$ in both (1) and (2) and setting them equal gives:

(4) $\frac{2a+3b}{2ab+b^{2}} = \frac{2b+3a}{2ba+a^{2}}$

Simplifying (4) such that there are no fractions and multiplying through will lead to:

(5) $2a^{3} + a^{2}b - b^{2}a - 2b^{3} = 0$

The following are some steps of algebraic simplification I made:

$ab(a-b) + 2(a^{3}-b^{3}) = 0$

$ab(a-b) + 2(a-b) (a^{2}+ab+b^{2}) = 0$

$(a-b) [ ab + 2 (a^{2}+ab+b^{2}) ] = 0$

(6) $(a-b)[2a^{2}+3ab+2b^{2}] = 0$

Obviously, based on (6), $a^{*} = b^{*}$ , where $(a^{*},b^{*})$ are the optimal pair of values. Additionally, you can find that the other term can only be simplified if you allow complex numbers. Since I am under the assumption we are looking for a pure real solution, I ignore that term.

Now plugging $a^{*} = b^{*}$ into (3), we obtain that:

$a^{*} = b^{*} = \sqrt[3]{\frac{\alpha}{2}}$

Given that $\alpha = 2000$ for our case, we can find that

$a^{*} = b^{*} = 10$

Plugging this into $F_{1}$ gives:

$F_{1}(a^{*},b^{*}) = \frac{1}{10} + \frac{1}{10} + \frac{1}{20} = \boxed{\frac{1}{4}}$

ab(a+b)=10x10x20 by comparing a=10 and b=10 i ticked 1/4 as the answer by this but i am not able to understand why 10 is the maximum value

since the function has one a kind of symmetry we can say that the max or min value (whichever exists) is achieved when a=b .

solving it we get a = b = 10 substituting the values we get that answer as 1/4

The expression can have minimum value, when a=b

Therefore, =>ab(a+b)=2000 =>a^2(2a)=2000 =>2a^3=2000 =>a=10

Simplifying the latter expression, =>1/a + 1/b + 1/a+b =>1/a + 1/a + 1/2a =>5/2a

Substituting a, we get =>5/20 = 1/4

$x\quad =\quad \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ (a+b) }$ $\Rightarrow x\quad =\quad \frac { (a+b)^{ 2 }+ab }{ ab(a+b) }$

From AM-GM, we have $(a+b)^{ 2 }\geq 4ab$

It is evident that for x to be minimum, $a = b$ $\Rightarrow a=10$

$\therefore \quad Least\quad value\quad of\quad x=\frac { 5a^{ 2 } }{ 2000 } =\quad \frac { 1 }{ 4 }$

I was told by a tutor in my school days that whenever a problem with symmetrical variables is given ,consider all of them equal for getting a maximum ora minimum....and it works always...put a=b and get to the answer in secomds orally..

I was wondering if you could use Lagrange Multipliers? I see that we have a function and a constraint in 2D, so applying the multipliers could definitely work? I ended up getting a symmetric equation in which the left side and the right side were the same except with a and b playing reversed roles.

As a result, I concluded that ONE solution is a = b (for sure, as if you reverse the roles of a and b and still have the same expression, surely a = b). From this I proved that a = b gives the minimum value of the sum, which turned out to be 1/4.

My question is if this is a valid approach? If not, can someone point out where my logic is flawed? Thank you so much!

These are cyclic expressions. By putting a=b=10 we get extremum. With these values, the required expression = .25. let us assume a=1. Solving we get the value of the required expression > .25. So, .25 is the minimum.

OR
$f(a,b)=\dfrac 1a+\dfrac 1b+\dfrac 1{a+b}=\dfrac {(a+b)^2+ab}{ab(a+b)}=\dfrac {(a+b)^2+ab}{2000}.\\ For ~extremum~ f_a'=0~ and ~f_b'=0.\\ \therefore~~~\dfrac{2(a+b)*1+b}{2000}=0~~~~~and~~~~~~\dfrac{2(a+b)*1+b}{2000}=0.\\ \implies~a=b,~~substituting~in~given ~equation~~~~~~~a^2*2a=2000,~~~~~~\implies~a=b=10.\\ So~f(10,10)=\dfrac1{10}+\dfrac1{10}+\dfrac1{20}=\dfrac 14.\\ Checking~by~2nd~derivative~this~ is~ the~ minimum.$