Prove: $x+y+z=1 ⇒ x^2 + y^2 + z^2 ≥ 1/3 \quad ∀x,y,z∈\mathbb R$

Question

Prove: $x+y+z=1 ⇒ x^2 + y^2 + z^2 ≥ \dfrac 13 \quad ∀x,y,z∈\mathbb R$

Can you help me to solve/prove this?

I'm trying to do so and still haven't found out the solution. I have been trying to transform the left side to look as the right side but I have no idea how to get there $≥$ sign (maybe it's all wrong).

I'll be glad if you help me.

no idea what it is. to be honest it looks more difficult than the task itself as i checked it on the internet :/ — user552297, Apr 14 '18 at 09:27
@user552297 A quick search gives this duplicate. I would be surprised if this question wasn't asked already; questions about Cauchy-Schwarz seem to appear a lot on this site. — Toby Mak, Apr 14 '18 at 09:30

score 1 · Answer 1 · answered Apr 14 '18 at 09:27

1

As the comment pointed, Cauchy-Schwartz inequality claims

$(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2$

then you substitute $a=1$, $b=1$, $c=1$,

$3(x^2+y^2+z^2) \geq (x+y+z)^2 =1^2=1$

S0 you will obtain the inequality.

answered Apr 14 '18 at 09:27

this_is_a_banana

152

Prove: $x+y+z=1 ⇒ x^2 + y^2 + z^2 ≥ 1/3 \quad ∀x,y,z∈\mathbb R$

1 Answers1