# If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.Find the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$

If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.

Find the minimum value of $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$.

I have proved it by using Lagrange's multiplier method.

Lagrange's multiplier method:

Let $f\displaystyle \equiv \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ and $\displaystyle g\equiv a+b+c+d-1=0$

Let $\displaystyle h\equiv \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\lambda(a+b+c+d-1)$

$\displaystyle \frac{\partial h}{\partial a}=\frac{-1}{a^2}-\lambda=0$

$\displaystyle \frac{\partial h}{\partial b}=\frac{-1}{b^2}-\lambda=0$

$\displaystyle \frac{\partial h}{\partial c}=\frac{-1}{c^2}-\lambda=0$

$\displaystyle \frac{\partial h}{\partial d}=\frac{-1}{d^2}-\lambda=0$

$\therefore a=b=c=d$

As $\displaystyle a+b+c+d=1\therefore a=b=c=d=\frac{1}{4}\therefore\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=16$

Is my way and answer correct?

Given $a+b+c+d = 1$ and We have to calculate $\bf{Minimum}$ of $\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$

where $a,b,c,d>0$

Using $\bf{A.M\geq H.M}$ Inequality

$$\displaystyle \frac{a+b+c+d}{4}\geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 16$$

And equality hold when $$a=b=c=d$$

Using $\bf{Cauchy -Schwartz}$ Inequality

$$\displaystyle \left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]\cdot \left[\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{c}}\right)^2+\left(\frac{1}{\sqrt{d}}\right)^2\right]\geq (1+1+1+1)^2$$

So We get $$\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 16$$

And equality hold when $$\frac{a}{1}=\frac{b}{1}=\frac{c}{1}=\frac{d}{1}$$