Let $a,b,c$ be positive real numbers such that $abc =1$

Question

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$. Also, state the condition for equality.

My Attempt, $a,b,c$ are real and positive numbers, then $$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$ $$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$ $$a^2+b^2+ c^2-2(a+b+c)+3\ge 0$$.

I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method.

@Dr.SonnhardGraubner It's not true without this assumption, take $a=b=c=1/2$ to see that. — Wojowu, Feb 02 '17 at 16:37
Another one here: How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? — Martin R, Feb 02 '17 at 18:35

score 7 · Answer 1 · edited Jun 12 '20 at 10:38

7

Note that from $$(a-1)^2+(b-1)^2+(c-1)^2 \ge 0$$we have $$a^2+b^2+ c^2 \ge 2(a+b+c)-3$$ By AM-GM $$a+b+c \ge 3 \sqrt[3]{abc}=3 \implies a+b+c-3 \ge 0$$ So $$a^2+b^2+ c^2 \ge 2(a+b+c)-3=a+b+c+(a+b+c-3) \ge a+b+c$$

EDIT

Here is a simple proof of AM-GM when $n=3$. We will prove $$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$ Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}, z=\sqrt[3]{c}$. The problem is equivalent to proving $$x^3+y^3+z^3-3xyz \ge 0$$ From here, we know $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ This can simplify to $$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) \ge 0$$ So we are done.

edited Jun 12 '20 at 10:38

Community

1

answered Feb 02 '17 at 16:16

S.C.B.

22,768

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@@ S.C.B, how did you use AM-GM? I didn't understand.. – pi-π Feb 02 '17 at 16:20
@NeWtoN I used AM-GM for when $n=3$. – S.C.B. Feb 02 '17 at 16:20
yeah but I only know when $y$ is the mean between $x$ and $z$. i.e, when, $$A.M=\frac {x+z}{2}$$ $$G.M=\sqrt {xz}$$. – pi-π Feb 02 '17 at 16:22
AM-GM says $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$. Multiply that inequality by $3$. – tylo Feb 02 '17 at 16:22
@NeWtoN What do you mean? – S.C.B. Feb 02 '17 at 16:24
@@S.C.B, I have edited my comment. Please do check. – pi-π Feb 02 '17 at 16:26
@NeWton Click the link I left in the answer. – S.C.B. Feb 02 '17 at 16:27
@NeWton Do you understand why AM-GM works for $n=3$? – S.C.B. Feb 02 '17 at 16:30
@@S.C.B, Yeah, I have said .. the link also provides me the concept when $A.M=\frac {x+y}{2}$ and $G.M=\sqrt {ab}$. But I did not get $A.M=\frac {x+y+z}{3}$. How is it please explain? Also, with G.M, too. – pi-π Feb 02 '17 at 16:32
@@S.C.B, No, I do not understand AM-GM, when $n=3$... – pi-π Feb 02 '17 at 16:33
@NeWtoN OK, I do not fully understand what you are saying, so I will edit it in my answer. – S.C.B. Feb 02 '17 at 16:35
@@S.C.B, I mean I.do not understand AM-GM for n=3. – pi-π Feb 02 '17 at 16:42
@NeWtoN Yes, so I added it in my answer. – S.C.B. Feb 02 '17 at 16:42
@Newton Do you understand? – S.C.B. Feb 02 '17 at 16:46
@S.C.B, I understood all except the last inequality, and how is $G.M=\sqrt [3] {abc}$? – pi-π Feb 02 '17 at 16:48
@Newton The last inequality just uses that all squares are positive, and GM is defined to be that way. – S.C.B. Feb 02 '17 at 16:48
@@S.C.B, It means if there are 4 terms G.M would have 4th root. – pi-π Feb 02 '17 at 16:50
@Newton Yes. It does. – S.C.B. Feb 02 '17 at 16:51
@Newton So do you understand it now? – S.C.B. Feb 02 '17 at 16:53
@@S.C.B, Yeah.. I am going through it again...Seems clear now. – pi-π Feb 02 '17 at 16:55
@Newton There is no need for double "@"s. You just need to add one. – S.C.B. Feb 02 '17 at 16:56

score 2 · Answer 2 · answered Feb 02 '17 at 20:27

2

By AM-GM $$6(a^2+b^2+c^2)=\sum_{cyc}(4a^2+b^2+c^2)\geq6\sum_{cyc}\sqrt[6]{a^8b^2c^2}=6(a+b+c)$$ and we are done!

answered Feb 02 '17 at 20:27

Michael Rozenberg

194,933

how did you start? Could you please explain. – pi-π Feb 03 '17 at 01:12
@NeWtoN It's just $6(a^2+b^2+c^2)=(4a^2+b^2+c^2)+(4b^2+c^2+a^2)+(4c^2+a^2+b^2)$ – Michael Rozenberg Feb 03 '17 at 04:45
Hello! Could you please explain your solution to me? – pi-π Apr 15 '17 at 15:36
@Ramanujan Yes, of course! By AM-GM $4a^2+b^2+c^2\geq6\sqrt[6]{(a^2)^4b^2c^2}=6a\sqrt[6]{a^2b^2c^2}=6a$ – Michael Rozenberg Apr 15 '17 at 15:42
How is that?.Could you explain more simply? – pi-π Apr 15 '17 at 15:53
@Ramanujan It's AM-GM for six variables. $x_1+x_2+...+x_n\geq n\sqrt[n]{x_1x_2...x_n}$ for positives $x_i$. – Michael Rozenberg Apr 15 '17 at 15:55
Where does the six variables come from? – pi-π Apr 15 '17 at 15:56
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@Ramanujan $4a^2+b^2+c^2=a^2+a^2+a^2+a^2+b^2+c^2$. – Michael Rozenberg Apr 15 '17 at 15:57
I didn't understamd from where and how you started and concluded the proof? – pi-π Apr 15 '17 at 15:59
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Let us continue this discussion in chat. – Michael Rozenberg Apr 15 '17 at 16:00

Let $a,b,c$ be positive real numbers such that $abc =1$

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