Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )
VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc
Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )
VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc
a)\(x^3+y^3+z^3-3xyz\\ \left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-\left[3xyz+3xy\left(x+y\right)\right]\\=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right] \\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+x^2-xy-xz-yz\right)\)
Ta có: \(S=a+b+c\left(1\right)\)
Thay \(\left(1\right)\)vào ta được:
\(\left(S-2b\right).\left(S-2c\right)=\left(a+b+c-2b\right).\)\(\left(a+b+c-2c\right)\)
\(=\left(a-b+c\right).\left(a+b-c\right)\)
\(=a^2+ab-ac-ba-b^2+bc+ca+cb-c^2\)
\(=a^2-b^2-c^2+2.bc\left(2\right)\)
Tương tự, ta được:
\(\left(S-2c\right).\left(S-2a\right)=b^2-c^2-a^2+2.ca\left(3\right)\)
\(\left(S-2a\right).\left(S-2b\right)=c^2-a^2-b^2+2.ab\left(4\right)\)
Từ \(\left(2\right);\left(3\right);\left(4\right)\Rightarrow\)Tổng bằng:
\(a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ca+c^2-a^2\)\(-b^2+2ab\)
\(=2ab+2bc+2ca-a^2-b^2-c^2\)
Vậy tổng trên \(=2ab+2bc+2ca-a^2-b^2-c^2.\)
\(8VT=4\left(a^2b+b^2c+c^2a+abc\right)\left(2ab^2+2bc^2+2ca^2+2abc\right)\le\left(a^2b+b^2c+c^2a+2ab^2+2bc^2+2ca^2+3abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+6abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+9abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left[\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\right]^2\)
\(\Rightarrow VT\le\frac{1}{512}\left[\left(a+2b\right)\left(4b+8c\right)\left(c+2a\right)\right]^2\)
\(\Rightarrow VT\le\frac{1}{512}\left(\frac{a+2b+4b+8c+c+2a}{3}\right)^6=\frac{1}{512}\left(a+2b+3c\right)^6=\frac{4^6}{512}=8\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;1;0\right)\)
\(1+a^2b^2=abc\left(a+b+c\right)+a^2b^2=ab\left(ab+bc+ca+c^2\right)=ab\left(a+c\right)\left(b+c\right)\)
\(1+b^2c^2=bc\left(a+b\right)\left(a+c\right)\) ; \(1+a^2c^2=ac\left(a+b\right)\left(b+c\right)\)
\(\Rightarrow Q=\frac{c^2\left(a+b\right)^2ab\left(a+c\right)\left(b+c\right)}{bc\left(a+b\right)\left(a+c\right)ac\left(a+b\right)\left(b+c\right)}=1\)
\(ab+bc+ca=2abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge\frac{2x-1}{2}\) \(\forall x:0< x< 2\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(2-x\right)^2\)
\(\Leftrightarrow9x^2-12x+4\ge0\)
\(\Leftrightarrow\left(3x-2\right)^2\ge0\) (luôn đúng)
Tương tự: \(\frac{y^3}{\left(2-y\right)^2}\ge\frac{2y-1}{2}\) ; \(\frac{z^3}{\left(2-z\right)^2}\ge\frac{2z-1}{2}\)
Cộng vế với vế: \(P\ge\frac{2\left(x+y+z\right)-3}{2}=\frac{4-3}{2}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\) hay \(a=b=c=\frac{3}{2}\)
