Can You Guess This ?

Using Cauchy Schwarz Inequality,

$\displaystyle \bigg({1} + {\frac{1}{2}} + {\frac{1}{3}} + {\frac{1}{4}} \bigg)({a}^{2} + 2{b}^{2} + 3{c}^{2} + 4{c}^{2}) \geq {(a + b + c + d)}^{2}$

$\frac{25}{12}({a}^{2} + 2{b}^{2} + 3{c}^{2} + 4{c}^{2}) \geq {5}^{2}$

$({a}^{2} + 2{b}^{2} + 3{c}^{2} + 4{c}^{2}) \geq 12$

Lagrange multipliers for $g(x_{1\dots n})=\sum_{k=1}^n{x_k-\frac{5}{n}}=0\,,$ $f(x_{1\dots n})=\sum_{k=1}^n{kx_k^2}$ give $\vec{\nabla} g=(1,\dots,1)\,,$ $\vec{\nabla}f=2(x_1,2x_2,\dots,nx_n)$ and, for $\vec{\nabla}f=\lambda\vec{\nabla}g$ , a single local extreme for $x_k=\frac{c}{k}$ .

We can find the value of $c$ by substituting back to the expression for $g=0$ , as $c=\frac{5}{H_n}$ , where $H_n=\sum_{k=1}^n\frac{1}{k}$ is the $n$ -th harmonic number. We can guess that the answer is a minimum, since there must be one (the value of our continuous function $f$ has a lower bound of 0, $\mathbb{R}_+^2$ is an open continuous space) and setting all but one of the variables $x_k$ to 0 points to a maximum instead (alternatively, by computing the Hessian of $f$ and showing that it's positive definite). The answer is therefore $\sum_{k=1}^n\frac{5^2}{H_n^2}k=\frac{25}{H_n}=12$ in case $n=4$ .

From Titu's Lemma; (a^2)/ 1+ (b^2)/ (1/2)+ (c^2)/ (1/3)+ (d^2)/ (1/4) will > or = (a+ b+ c+ d)^ 2/ (1+ (1/2)+ (1/3)+ (1/4)) but (a+ b+ c+ d)= 5 so, a^2+ 2 b^2+ 3 c^2+ 4*d^2 > or = (5)^2/ (25/ 12) = 12

We can also use weighted mean inequality

i too used titu lemma