Well, $ (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) = 1 \rightarrow 2(ab+bc+ca) = 1 - (a^2+b^2+c^2) \rightarrow 4(ab+bc+ca) = 2 - 2(a^2+b^2+c^2)$ So we have to prove $ a^2+b^2+c^2 \geq 2 - 2(a^2+b^2+c^2) - 1 \rightarrow 3(a^2+b^2+c^2) \geq 1 \rightarrow a^2+b^2+c^2 \geq \frac{1}{3}$ But for QM-AM inequality we have $ \sqrt{\frac{a^2+b^2+c^2}{3}}\geq \frac{a+b+c}{3} \rightarrow \frac{a^2+b^2+c^2}{3} \geq \left(\frac{1}{3}\right)^2 \rightarrow {a^2+b^2+c^2} \geq \frac{1}{3}$ as wanted.

Edit: Fixed the typo pointed out by mathwizarddude. Almost five years later

DreamTeam wrote: Let $ a,b,c\geq 0$ such that $ a + b + c = 1$. Prove that: $ a^2 + b^2 + c^2\geq 4(ab + bc + ca) - 1$ $ a^2 + b^2 + c^2\geq 4(ab + bc + ca) - 1$ $ \leftrightarrow (a+b+c)^2 -2(ab+bc+ca) \ge\ 4(ab + bc + ca) - 1$ $ \leftrightarrow 1 \ge\ 3(ab+bc+ca)$ which is obvious We have done!

It's easy to see that: $ a^2+b^2+c^2=(a+b+c)(a^2+b^2+c^2)\geq 9abc$(AM-GM inequality). We can prove a stronger inequality: \[ 9abc\geq 4(ab+bc+ca)-1.\] After normalization the inequality becomes: \[ \frac{9abc}{a+b+c}\geq 4(ab+bc+ca)-(a+b+c)^2\] \[ \Longleftrightarrow (a+b+c)^3+9abc\geq 4(a+b+c)(ab+bc+ca)\] \[ \Longleftrightarrow a^3+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)\] \[ \Longleftrightarrow\displaystyle\sum_{cyc}a(a-b)(a-c)\geq 0\] which is Schur's Inequality for $ \boxed{r=1}.$

By Power Mean Inequality: $ \left(\frac{a^2+b^2+c^2}3\right)^{\frac12}\ge\frac{a+b+c}3=\frac13$, and hence $ 3(a^2+b^2+c^2)\ge1$. So $ a^2+b^2+c^2\ge 2-2(a^2+b^2+c^2)-1=2(1-a^2+b^2+c^2)-1=2((a+b+c)^2-a^2+b^2+c^2)-1=2(2ab+2bc+2ca)-1=4(ab+bc+ca)-1$. Q.E.D.

Another way(Using Mixing varialble) Let $ f\left( {a,b,c} \right) = a^2 + b^2 + c^2 - 4(ab + bc + ac) + 1$ so $ f\left( {a,\frac{{b + c}}{2},\frac{{b + c}}{2}} \right) = a^2 + (\frac{{b + c}}{2})^2 + (\frac{{b + c}}{2})^2 - 4(ab + (\frac{{b + c}}{2})^2 + ac) + 1$ And $ f(a,b,c) - f\left( {a,\frac{{b + c}}{2},\frac{{b + c}}{2}} \right) \ge 0$ So we just have to prove that $ f\left( {a,\frac{{b + c}}{2},\frac{{b + c}}{2}} \right) \ge 0$ which is obvious.

$ a^2+b^2+c^2\geq ab+bc+ca$ by rearrangement inequality/AM-GM/Anything really so $ 2(a^2+b^2+c^2) + 2(ab+bc+ca)\geq 4(ab+bc+ca)$ $ a^2+b^2+c^2 + (a+b+c)^2 = a^2+b^2+c^2 +1 \geq 4(ab+bc+ca)$ $ a^2+b^2+c^2 \geq 4(ab+bc+ca)-1$

TZF wrote: Expand and rearrange: $ 2a^2 + 2b^2 + 2c^2 - 2a - 2b - 2c \geq 0$ I think it should have been $ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac$, but anyways, nice solution.

let $ a^2+b^2+c^2 \geq 4(ab+bc+ca) - 1$ be denoted by (P). Then, $ a+b+c=1 \Rightarrow (a+b+c)^2=1;$ hence: $ (P) \Leftrightarrow a^2+b^2+c^2 \geq 4(ab+bc+ca) - a^2 - b^2 - c^2 - 2(ab+bc+ca)$ $ \Leftrightarrow a^2+b^2+c^2 \geq ab+bc+ca$, which is true by AM-GM. Hence the result.

DreamTeam wrote: Let $ a,b,c\geq 0$ such that $ a + b + c = 1$. Prove that: $ a^2 + b^2 + c^2\geq 4(ab + bc + ca) - 1$

It is very is inequality.

DreamTeam wrote: Let $ a,b,c\geq 0$ such that $ a + b + c = 1$. Prove that: $ a^2 + b^2 + c^2\geq 4(ab + bc + ca) - 1$

Mewto55555 wrote: DreamTeam wrote: Let $ a,b,c\geq 0$ such that $ a + b + c = 1$. Prove that: $ a^2 + b^2 + c^2\geq 4(ab + bc + ca) - 1$

dont u think this is overkill metwo5555? i mean it is very simple then why go for a method like langrange multipliers..

Well since we have the constraint of a+b+c=1, then this just means all the partials are equal, and its really easy to find the critical points.

$ a^2+b^2+c^2>=ab+bc+ca $ $ 2(a^2+b^2+c^2)>=2(ab+bc+ca) $ $ a^2+b^2+c^2+(a+b+c)^2>=4(ab+bc+ca) $ $ a^2+b^2+c^2+1>=4(ab+bc+ca) $ $ a^2+b^2+c^2>=4(ab+bc+ca)-1 $

somebody wrote: $ a^2+b^2+c^2%Error. "geqab" is a bad command. +bc+ca $ $ 2(a^2+b^2+c^2)>=2(ab+bc+ca) $ $ a^2+b^2+c^2+(a+b+c)^2>=4(ab+bc+ca) $ $ a^2+b^2+c^2+1>=4(ab+bc+ca) $ $ a^2+b^2+c^2>=4(ab+bc+ca)-1 $

$ a^2+b^2+c^2\geq ab+bc+ca $ $ 2(a^2+b^2+c^2)\geq 2(ab+bc+ca) $ $ a^2+b^2+c^2+(a+b+c)^2\geq 4(ab+bc+ca) $ $ a^2+b^2+c^2+1\geq 4(ab+bc+ca) $ $ a^2+b^2+c^2\geq 4(ab+bc+ca)-1 $ QED

Since $(a+b+c)^2=1$, $a^2+b^2+c^2 \geq 4(ab+bc+ca)-1 \Leftrightarrow a^2+b^2+c^2 \geq ab+bc+ca$ This is just Rearrangement inequality