if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

Question

This is supposed to be an application of AM-GM inequality.

if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$

First of all,

$a^2+b^2+c^2\ge 3$

by a direct application of AM-GM.Also,we have

$a^2+b^2+c^2\ge ab+bc+ca$

Next,we consider the expression

$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$

but that hardly helps.I know that

$3(a^2+b^2+c^2)\ge (a+b+c)^2$

From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,

$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$

$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$ A hint will be appreciated at this point.

Does this answer your question? Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$. — xskxzr, Feb 18 '21 at 01:34

score 15 · Accepted Answer · edited Mar 23 '21 at 19:58

15

Notice that for all real $a,b,c$,

$$(a - 1)^2 + (b-1)^2 + (c - 1)^2 \ge 0.$$ $$a^2 + b^2 + c^2 - 2a - 2b - 2c + 3 \ge 0.$$ $$a^2 + b^2 + c^2 \ge -3 + (a + b + c) + (a + b + c).$$

But by AM-GM, $a + b + c \ge 3\sqrt[3]{abc} = 3$. So, $$a^2 + b^2 + c^2 \ge -3 + 3 + (a + b + c).$$ $$a^2 + b^2 + c^2 \ge a + b + c.$$

Equality is attained when $a = b = c = 1$.

(I really wish to put everything after the first line as a spoiler so that this answer becomes a hint, but I don't know how D:)

edited Mar 23 '21 at 19:58

amWhy

209,954

answered Apr 05 '14 at 08:29

Yiyuan Lee

14,435

1

A spoiler is generated by starting your line with >! – celtschk Apr 05 '14 at 09:21
Nice answer.May I know the motivation behind considering $(a-1)^2+(b-1)^2+(c-1)^2$? – rah4927 Apr 05 '14 at 15:31
I thought about how one can use $(a - b)^2 + (b - c)^2 + (c - a)^2$ to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$, and played around to get that :) – Yiyuan Lee Apr 05 '14 at 15:38
The AM-GM inequality is only generally applicable for non-negative real numbers. – user236182 Jul 25 '17 at 14:22
@user236182 In the question it is mentioned that AM-GM is allowed for $a + b + c$. – Yiyuan Lee Jul 26 '17 at 05:26

score 7 · Answer 2 · answered Jul 25 '17 at 14:11

Use the fact that $$a^2+a^2+a^2+b+c \geq5\sqrt[5]{a^2.a^2.a^2.b.c}=5\sqrt[5]{a^6bc}=5a$$ since $abc=1$

Similarly, we get two more results, and adding them we get: $$3(a^2+b^2+c^2)+2(a+b+c) \geq 5(a+b+c) $$ which gives us our result. $$a^2+b^2+c^2 \geq a+b+c$$ $$Q.E.D.$$

score 6 · Answer 3 · answered Apr 05 '14 at 09:17

Here is an exotic solution based on geometry.

Let $\mathcal{M}$ and $\mathcal{S}$ be surfaces defined by

\begin{align*} \mathcal{M} : abc = 1 \quad \text{and} \quad \mathcal{S} : a^{2} + b^{2} + c^{2} = a + b + c. \end{align*}

Then we have the following observations:

$\mathcal{M}$ lies outside the sphere of radius $\sqrt{3}$. Indeed, if $X = (a, b, c) \in \mathcal{M}$, then the square-distance from the origin $O$ satisfies $$ \overline{OX}^{2} = a^{2} + b^{2} + c^{2} \geq 3\sqrt[3]{a^{2}b^{2}c^{2}} = 3. $$
$\mathcal{S}$ is contained in the sphere of radius $\sqrt{3}$. Indeed, $\mathcal{S}$ is the sphere of radius $\frac{\sqrt{3}}{2}$ centered at the point $P = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$, hence if $X = (a, b, c) \in \mathcal{S}$ then by the triangle inequality

$$ \overline{OX} \leq \overline{OP} + \overline{PX} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}.$$

Combining two facts, we find that $\mathcal{M}$ lies outside of $\mathcal{S}$ (possible except for the tangent point). Therefore

\begin{align*} abc = 1 & \quad \Longrightarrow \quad (a, b, c) \in \mathcal{M} \\ & \quad \Longrightarrow \quad (a, b, c) \text{ lies outside } \mathcal{S} \\ & \quad \Longrightarrow \quad \left(a - \tfrac{1}{2}\right)^{2} + \left(b - \tfrac{1}{2}\right)^{2} + \left(c - \tfrac{1}{2}\right)^{2} \geq \tfrac{3}{4} \\ & \quad \Longrightarrow \quad a^{2} + b^{2} + c^{2} \geq a + b + c. \end{align*}

M lives outside the sphere of which center? – Math Mar 14 '23 at 16:22 — Math, Mar 14 '23 at 16:22

score 2 · Answer 4 · answered Apr 08 '17 at 11:10

2

Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, our inequality it's just Muirhead.

answered Apr 08 '17 at 11:10

Michael Rozenberg

194,933

Love how all polynomial ineqs are muirhead-able or schur-able :v – rah4927 Apr 08 '17 at 19:13

user133281 · Answer 5 · 2014-04-05T10:06:43.823

2

Alternatively, $a^2+a^2+a^2+a^2+b^2+c^2 \geq 6 \sqrt[6]{a^8b^2c^2} = 6 \sqrt[6]{a^6} = 6|a| \geq 6a$ by AM-GM. Adding the analogous inequalities $a^2+4b^2+c^2 \geq 6b$ and $a^2+b^2+4c^2 \geq 6c$ gives the result.

edited Apr 05 '14 at 10:06

answered Apr 05 '14 at 08:46

user133281

16,073

But what if $a<0$? – celtschk Apr 05 '14 at 09:23
Notice that $a^2$, $b^2$ and $c^2$ are non-negative, thus applying AM-GM is allowed. We have $\sqrt[6]{a^6} = |a| \geq a$. – user133281 Apr 05 '14 at 10:06
With that addition, the proof is correct (and actually quite nice). Note that I had already upvoted you before I thought about the negative number problem, and forgot to unupvote when I noticed it, so don't be surprised that I don't upvote after this comment. – celtschk Apr 05 '14 at 10:17
Nice.But may I know the motivation behind considering $4a^2+b^2+c^2$? – rah4927 Apr 05 '14 at 15:25

score 1 · Answer 6 · answered Aug 05 '20 at 08:21

1

$GM\le AM\le RMS\implies GM\times AM\le RMS^2\implies a+b+c\le a^2+b^2+c^2$.

answered Aug 05 '20 at 08:21

Chrystomath

10,798

if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

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