# 30 BÀI TOÁN BBT ĐĐNG THHC

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.

Bạn đã xem: =1/3(A+B+C)^2">Chứng Minh A^2+B^2+C^2>=1/3(A+B+C)^2

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

Bạn đang xem: 30 bài toán bbt đđng thhc  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

Xem thêm: Lg Tai Nghe Lg Bluetooth Tone Free Hbs Fn4 Chính Hãng, Tai Nghe Lg Tone Free Hbs  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.\$\$

Xem thêm: Truyện Hũ Bạc Của Người Cha Tiếng Việt 3, Hũ Bạc Của Người Cha (Trang 121) 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*