Cauchy-Schwarz is pretty helpful

The problem is a very easy application of AM-HM. We know that for the fout positive real numbers $a,b,c,d$ , By AM-HM

$\frac{a+b+c+d}{4} \geq \frac{4}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$ .

This this is very simple that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})(a+b+c+d)≥16$

But since, $a+b+c+d=1$ , we get $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})≥16$ Therefore, the minimum value is 16

$\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\\\stackrel{\text{Cauchy-Schwarz}}\ge (1+1+1+1)^2=16$ .

Since $16$ can be reached when $\displaystyle\cfrac{a}{\cfrac{1}{a}}=\cfrac{b}{\cfrac{1}{b}}=\cfrac{c}{\cfrac{1}{c}}=\cfrac{d}{\cfrac{1}{d}}\implies a^2=b^2=c^2=d^2\stackrel{a,b,c,d>0}{\implies} a=b=c=d=\frac{1}{4}\text{,}$ the answer is $\boxed{16}$ .

If $a + b + c + d = 1$ , then $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} = 1 * (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} ) =$ $(a + b + c + d) * (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d})$ But $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} = (\frac {1}{ \sqrt {a}})^2 + (\frac {1}{ \sqrt {b}})^2 + (\frac {1}{ \sqrt {c}})^2 + (\frac {1}{ \sqrt {d}})^2$ and $(a + b + c + d) = (\sqrt {a})^2 + (\sqrt {b})^2 + (\sqrt {c})^2 + (\sqrt {d})^2$ So $(a + b + c + d) * (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d}) =$ $((\sqrt {a})^2 + (\sqrt {b})^2 + (\sqrt {c})^2 + (\sqrt {d})^2) *((\frac {1}{ \sqrt {a}})^2 + (\frac {1}{ \sqrt {b}})^2 + (\frac {1}{ \sqrt {c}})^2 + (\frac {1}{ \sqrt {d}})^2)$ By Cauchy-Schwarz Inequality,

$\geq (\sqrt {a} * \frac {1}{ \sqrt {a}} + \sqrt {b} * \frac {1}{ \sqrt {b}} + \sqrt {c} * \frac {1}{ \sqrt {c}} + \sqrt {d} * \frac {1}{ \sqrt {d}})^2 =$ $(1 + 1 + 1 + 1)^2 = (4)^2 = \boxed{16}$

simple...put a=b=c=d=1/4...So, 1/a+1/b+1/c+1/d=16

According to Bunyakovsky inequality: $\frac{a+b+c+d}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} \ge (1+1+1+1)^2=4^2=16$ $\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \ge 16$ $MIN((\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=\boxed{16}$ "=" happens when a=b=c=d=1/4

We can get it by directly applying Titu's lemma

$a + b + c +d = 1$

According to Cauchy–Schwarz inequality:

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) \left(a+b+c+d\right) \ge \left(\frac{\sqrt{a}}{\sqrt{a}} +\frac{\sqrt{b}}{\sqrt{b}} +\frac{\sqrt{c}}{\sqrt{c}} +\frac{\sqrt{d}}{\sqrt{d}}\right)^2$

Therefore:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \ge 16$

By Cauchy-Schwarz inequality, [ $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ + $\frac{1}{d}$ ] (a+b+c+d) $\geq$ { $\sqrt{a/a}$ + $\sqrt{b/b}$ + $\sqrt{c/c}$ + $\sqrt{d/d}$ } $\times$ { $\sqrt{a/a}$ + $\sqrt{b/b}$ + $\sqrt{c/c}$ + $\sqrt{d/d}$ } Which becomes $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ + $\frac{1}{d}$ $\geq$ 4 $\times$ 4 Therefore the minimum value is 16

I know of the obvious solution, but please point out where i am going wrong in this approach Using Holder's Inequality: [ a^-1 + b^-1 + c^-1 + d^-1 ] [ 1^-1 +1^-1 +1^-1 +1^-1 ] >= [ a+ b+ c+ d ]^-1 => Required >= 1/4

Easy one. a+b+c+d=1, this will be called eq(1), the minimun its obtained when all numbers are the same, then suppose a=b=c=d=k, replacing on eq(1) we have k=1/4. Then 1/a+1/b+1/c+1/d became 4*(1/1/4)=16.

(a+b+c+d) *((1/a +(1/b) +(1/c)+(1/d)) >_ (1 +1 +1 +1)^2 or 16 then we realize that the solution must be the same or 1/4

AM-GM also works easily. Applying it twice, we can get $\frac{a+b+c+d}{4} \ge \sqrt[4]{abcd}$ $\text{and}$ $\frac{\frac 1a + \frac 1b + \frac 1c + \frac 1d}{4} \ge \sqrt[4]{\frac{1}{abcd}}.$ Multiplying both equations together, we get $(a+b+c+d)\left(\frac 1a + \frac 1b + \frac 1c + \frac 1d\right) \ge \sqrt[4]{1}\times 16.$ Noting that $a+b+c+d = 1$ , we get that $\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \boxed{16}.$