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đăng từng câu 1 thôi, nhiều nhất là 3 câu/ 1 lần hỏi vì đâu có giới hạn số lần hỏi
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
Ta có: \(a^3+b^3+c^3-3abc=0\) \(\Leftrightarrow a+b+c=0\) hoặc a = b = c
theo gt thi a + b + c \(\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow N=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
\(ab=x;bc=y;ac=z\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)
Tự full nhé?
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Lời giải:
\((a^2+b^2)^3=(a^3+b^3)^2\)
\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6\)
\(\Rightarrow 3a^2b^2(a^2+b^2)=2a^3b^3\)
\(\Rightarrow 3(a^2+b^2)=2ab\) (chia cả 2 vế cho \(a^2b^2\neq 0\) )
\(\Rightarrow 2(a^2+b^2)=-(a^2+b^2)+2ab=-(a-b)^2(*)\)
Với mọi \(a,b\neq 0\) thì \(2(a^2+b^2)>0; -(a-b)^2\leq 0\), do đó $(*)$ vô lý
Vậy không tồn tại $a,b$ thỏa mãn đkđb, kéo theo không tìm được giá trị của $A$
Bổ sung ĐK không âm nha bạn 
Áp dụng BĐT Cô - Si : x2 + y2 ≥ 2xy ( x > 0 ; y > 0)
Ta có :
a2 + b2 ≥ 2ab ( a > 0 ; b > 0) ( 1 )
b2 + c2 ≥ 2bc ( b > 0 ; c > 0) ( 2)
c2 + a2 ≥ 2ac ( c > 0 ; a > 0) ( 3 )
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta có :
2( a2 + b2 + c2 ) ≥ 2( ab + bc + ac)
⇔ 3( a2 + b2 + c2) ≥ a2 + b2 + c2 + 2ab + 2bc + 2ac
⇔ a2 + b2 + c2 ≥ \(\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{9}{\dfrac{4}{3}}=\dfrac{3}{4}\)
Dấu " = " xảy ra khi : a = b = c = \(\dfrac{1}{2}\)
Ta có:
(a+b+c)2=a2+b2+c2
a2+b2+c2+2ab+2ac+2bc=a2+b2+c2
2(ab+bc+ca)=0
ab+bc+ca=0
Ta có:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\dfrac{3}{abc}\)
\(\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=3\)
\(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\left(ab+bc\right)^3-3ab^2c\left(ab+bc\right)+a^3c^3-3a^2b^2c^2=0\)
\(\left(ab+bc+ca\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ca\right)-3ab^2c\left(-ac\right)-3a^2b^2c^2=0\)
\(0+3a^2b^2c^2-3a^2b^2c^2+0=0\)
0=0(luôn đúng)
Vậy BĐT được chứng minh
Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)-a^2-b^2-c^2=0\)
\(\Rightarrow ab+bc+ca=0\)
\(\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Chia cả 2 vế cho \(a^3b^3c^3\) , ta có :
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\left(đpcm\right)\)
a: \(\left(ax-by\right)^2+\left(bx+ay\right)^2\)
\(=a^2x^2-2axby+b^2y^2+b^2x^2+2abxy+a^2y^2\)
\(=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(a^2+b^2\right)\)
c: \(a^2+2ab+b^2-c^2\)
\(=\left(a+b\right)^2-c^2\)
\(=\left(a+b+c\right)\left(a+b-c\right)\)
\(=4m\cdot\left(4m-2c\right)\)
\(=16m^2-8mc\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2-c^2=2bc\\b^2-a^2-c^2=2ac\\c^2-a^2-b^2=2ab\end{matrix}\right.\)
\(\Leftrightarrow Q=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
Trường hợp 2: a=b=c
\(\Leftrightarrow Q=\dfrac{a^2}{a^2-a^2-a^2}+\dfrac{b^2}{b^2-b^2-b^2}+\dfrac{c^2}{c^2-c^2-c^2}\)
\(\Leftrightarrow Q=\dfrac{a^2}{-2a^2}+\dfrac{b^2}{-2b^2}+\dfrac{c^2}{-2c^2}=\dfrac{-1}{2}+\dfrac{-1}{2}+\dfrac{-1}{2}=\dfrac{-3}{2}\)