Let $x,y,z,w $ be real numbers such that $x+2y+3z+4w=1$. Find the minimum of $x^2+y^2+z^2+w^2+(x+y+z+w)^2$.
Problem
Source: China Sijiazhuang , Aug 2014
Tags: inequalities, inequalities proposed
12.08.2014 05:22
We use Cauchy's inequality: \[ [(-1)^2 +0^2+1^2+2^2+2^2][x^2+y^2+z^2+w^2+(x+y+z+w)^2]\\ \geq(-x+z+2w+2x+2y+2z+2w)^2\\ =(x+2y+3z+4w)^2\\ =1 \] so \[ 10( x^2+y^2+z^2+w^2+(x+y+z+w)^2)\geq1 \] the answer is $\frac{1}{10}$, and equality holds when \[ x=-0.1\\ y=0\\ z=0.1\\ w=0.2 \]
12.08.2014 09:02
Very nice. Thank you very much.
23.08.2015 04:51
Those are possible number? x = -0.1y = 0z = 0.1w = 0.2 0z = 0.2 ? I have no idea.
31.01.2017 10:59
fanzhuyifan wrote: We use Cauchy's inequality: \[ [(-1)^2 +0^2+1^2+2^2+2^2][x^2+y^2+z^2+w^2+(x+y+z+w)^2]\\ \geq(-x+z+2w+2x+2y+2z+2w)^2\\ =(x+2y+3z+4w)^2\\ =1 \]so \[ 10( x^2+y^2+z^2+w^2+(x+y+z+w)^2)\geq1 \]the answer is $\frac{1}{10}$, and equality holds when \[ x=-0.1\\ y=0\\ z=0.1\\ w=0.2 \] thanks a lot
14.06.2022 11:47
actually This problem is very easy if you use partial derivative but I think it can't be a solution written on the test paper