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I am studying for math competitions but I usually have trouble with problems of the sort:

Let $a,b,c,d$ sum to $1$ and they are all positive real numbers. Find $(a,b,c,d)$ given that $1/a+4/b+9/c+16/d$ is minimum and what is this minimum value.

How do I approach the problem? Are Lagrange multipliers useful? Are there any other techniques? What should I know for olympiad level inequalities?

Edit: Most of these can be solved by the Lagrange method but not efficiently. Any substitutes for this?

2 Answers 2

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Lagrange multipliers are useful, although they are not strictly necessary for this problem. ( because the expression $a+b+c+d=1$ does not mix variables)

Let us minimize the expression $a+x/b$ subject to $a+b=2w$ (here $1

Since $a+b=2w$ we can rewrite $a$ and $b$ as $w-\alpha$ and $w+\alpha$.

So now our expression is $\frac{1}{w-\alpha}+\frac{x}{w+\alpha}$. To minimize the expression take derivative and get $\frac{1}{(w-\alpha)^2}-\frac{x}{(w+\alpha)^2}$ which is clearly solved by $\frac{w+\alpha}{w-\alpha}=\sqrt x$

So if $a=z$ we must have $b=2z,c=3z,d=4z$. So we have $z=\frac{1}{10}$.

Hence the answer is: $(\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10})$

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    That's pretty good! Where did you get the w and x/b+a inspiration though? It doesn't seem like a natural step. – 2017-01-08
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    The idea is to change $a$ and $b$ into a single variable (subject to the fixed sum) – 2017-01-08
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If you're talking about Olympiads or related competitions, I think calculus is actually forbidden or at least highly discouraged (if you do not state the theorem exactly or do not verify the assumptions many points will fly away).

In these competitions there is often a proof by Mean inequalities with an ad hoc application. In this case by the inequality between the Arithmetic and the Harmonic Mean $$\frac{a\!+\!\left(\frac b2\!+\!\frac b2\right)\!+\!\left(\frac c3\!+\!\frac c3\!+\!\frac c3\right)\!+\!\left(\frac d4\!+\!\frac d4\!+\!\frac d4\!+\!\frac d4\right)}{10}\geq \frac{10}{\tfrac1a\!+\!\left(\tfrac2b\!+\!\tfrac2b\right)\!+\!\left(\tfrac3c\!+\!\tfrac3c\!+\!\tfrac3c\right)\!+\!\left(\tfrac4d\!+\!\tfrac4d\!+\!\tfrac4d\!+\!\tfrac4d\right)}$$ from which $$\frac1a+\frac4b+\frac9c+\frac{16}{d}\geq 100$$ and the equality is achieved, as in every Mean inequality, when all the terms are equal, in this case $$a=\frac b2=\frac c3=\frac d4.$$ From $a+b+c+d=1$ you can find the exact values $a=\frac{1}{10},b=\frac{2}{10},c=\frac{3}{10},d=\frac{4}{10}$.

EDIT: You can also use Cauchy-Schwartz: $$\left(\frac{1}{\sqrt a}\sqrt a+\frac{2}{\sqrt b}\sqrt b+\frac{3}{\sqrt c} \sqrt c+\frac{4}{\sqrt d}\sqrt d\right)^2\leq\left(\frac1a+\frac4b+\frac9c+\frac{16}{d}\right)\left(a+b+c+d\right)$$