Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 ge a + b + c$. (4 answers) This is supposed to lớn be an application of AM-GM inequality.

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

First of all,

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

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

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

Next,we consider the expression

$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abcle a+b+c+a^2+b^2+c^2+1$

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 lớn 9.But I can"t see how that can be used here.I know that we can get $a+b+cge 3$ using AM-GM but that does not take me a step closer to lớn finding the solution(from what I can understand).Also,

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

algebra-precalculus inequality symmetric-polynomials a.m.-g.m.-inequality muirhead-inequality nói qua Cite Follow edited Oct 19 "18 at 3:00

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Michael Rozenberg 180k2828 gold badges142142 silver badges247247 bronze badges asked Apr 5 "14 at 8:22

rah4927rah4927 3,49422 gold badges1818 silver badges4949 bronze badges $endgroup$ 1 địa chỉ a phản hồi |

6 Answers 6

Active Oldest Votes 14 $egingroup$ 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, $a + b + c ge 3sqrtabc = 3$. 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 khổng lồ put everything after the first line as a spoiler so that this answer becomes a hint, but I don"t know how D:)

chia sẻ Cite Follow edited Mar 23 at 19:58

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amWhy 201k135135 gold badges257257 silver badges477477 bronze badges answered Apr 5 "14 at 8:29

Yiyuan LeeYiyuan Lee 14k44 gold badges3737 silver badges6666 bronze badges $endgroup$ 5 địa chỉ cửa hàng a phản hồi | 6 $egingroup$ Use the fact that $$a^2+a^2+a^2+b+c geq5sqrta^2.a^2.a^2.b.c=5sqrta^6bc=5a$$ since $abc=1$

Similarly, we get two more results, & 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.$$a^2+b^2+c^2 geq a+b+c$$$$Q.E.D.$$

tóm tắt Cite Follow answered Jul 25 "17 at 14:11

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Aditya KhurmiAditya Khurmi 6111 silver badge11 bronze badge $endgroup$ địa chỉ cửa hàng a phản hồi | 4 $egingroup$ Here is an exotic solution based on geometry.

Let $goodsmart.com.vncalM$ và $goodsmart.com.vncalS$ be surfaces defined by

eginalign*goodsmart.com.vncalM : abc = 1quad extand quadgoodsmart.com.vncalS : a^2 + b^2 + c^2 = a + b + c.endalign*

Then we have the following observations:

$goodsmart.com.vncalM$ lies outside the sphere of radius $sqrt3$. Indeed, if $X = (a, b, c) in goodsmart.com.vncalM$, then the square-distance from the origin $O$ satisfies$$ overlineOX^2 = a^2 + b^2 + c^2 geq 3sqrta^2b^2c^2 = 3. $$

$goodsmart.com.vncalS$ is contained in the sphere of radius $sqrt3$. Indeed, $goodsmart.com.vncalS$ is the sphere of radius $fracsqrt32$ centered at the point $P = (frac12, frac12, frac12)$, hence if $X = (a, b, c) in goodsmart.com.vncalS$ then by the triangle inequality

$$ overlineOX leq overlineOP + overlinePX = fracsqrt32 + fracsqrt32 = sqrt3.$$

Combining two facts, we find that $goodsmart.com.vncalM$ lies outside of $goodsmart.com.vncalS$ (possible except for the tangent point). Therefore

eginalign*abc = 1& quad Longrightarrow quad (a, b, c) in goodsmart.com.vncalM \& quad Longrightarrow quad (a, b, c) ext lies outside goodsmart.com.vncalS \& quad Longrightarrow quad left(a - frac12 ight)^2 + left(b - frac12 ight)^2 + left(c - frac12 ight)^2 geq frac34 \& quad Longrightarrow quad a^2 + b^2 + c^2 geq a + b + c.endalign*