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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.

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    why do we need $$abc=1$$? – 2017-02-02
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    So that $\sqrt[3]{abc} = 1$, see answer below. – 2017-02-02
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    @NeWton We need it so that $$3 \sqrt[3]{abc}=3$$ – 2017-02-02
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    @Dr.SonnhardGraubner It's not true without this assumption, take $a=b=c=1/2$ to see that. – 2017-02-02
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    Another one here: [How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$?](http://math.stackexchange.com/questions/501106/how-to-prove-abc-le-a2b2c2-if-abc-1) – 2017-02-02
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    @AlbertEinstein Why did you unaccept my answer? – 2017-04-29

2 Answers 2

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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.

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    @@ S.C.B, how did you use AM-GM? I didn't understand.. – 2017-02-02
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    @NeWtoN I used AM-GM for when $n=3$. – 2017-02-02
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    yeah but I only know when $y$ is the mean between $x$ and $z$. i.e, when, $$A.M=\frac {x+z}{2}$$ $$G.M=\sqrt {xz}$$. – 2017-02-02
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    AM-GM says $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$. Multiply that inequality by $3$. – 2017-02-02
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    @NeWtoN What do you mean? – 2017-02-02
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    @@S.C.B, I have edited my comment. Please do check. – 2017-02-02
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    @NeWton Click the link I left in the answer. – 2017-02-02
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    @NeWton Do you understand why AM-GM works for $n=3$? – 2017-02-02
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    @@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. – 2017-02-02
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    @@S.C.B, No, I do not understand AM-GM, when $n=3$... – 2017-02-02
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    @NeWtoN OK, I do not fully understand what you are saying, so I will edit it in my answer. – 2017-02-02
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    @@S.C.B, I mean I.do not understand AM-GM for n=3. – 2017-02-02
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    @NeWtoN Yes, so I added it in my answer. – 2017-02-02
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    @Newton Do you understand? – 2017-02-02
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    @S.C.B, I understood all except the last inequality, and how is $G.M=\sqrt [3] {abc}$? – 2017-02-02
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    @Newton The last inequality just uses that all squares are positive, and GM is defined to be that way. – 2017-02-02
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    @@S.C.B, It means if there are 4 terms G.M would have 4th root. – 2017-02-02
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    @Newton Yes. It does. – 2017-02-02
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    @Newton So do you understand it now? – 2017-02-02
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    @@S.C.B, Yeah.. I am going through it again...Seems clear now. – 2017-02-02
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    @Newton There is no need for double "@"s. You just need to add one. – 2017-02-02
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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!

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    how did you start? Could you please explain. – 2017-02-03
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    @NeWtoN It's just $6(a^2+b^2+c^2)=(4a^2+b^2+c^2)+(4b^2+c^2+a^2)+(4c^2+a^2+b^2)$ – 2017-02-03
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    Hello! Could you please explain your solution to me? – 2017-04-15
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    @Ramanujan Yes, of course! By AM-GM $4a^2+b^2+c^2\geq6\sqrt[6]{(a^2)^4b^2c^2}=6a\sqrt[6]{a^2b^2c^2}=6a$ – 2017-04-15
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    How is that?.Could you explain more simply? – 2017-04-15
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    @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 $x_i$. – 2017-04-15
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    Where does the six variables come from? – 2017-04-15
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    @Ramanujan $4a^2+b^2+c^2=a^2+a^2+a^2+a^2+b^2+c^2$. – 2017-04-15
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    I didn't understamd from where and how you started and concluded the proof? – 2017-04-15
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/57157/discussion-between-michael-rozenberg-and-ramanujan). – 2017-04-15