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Theo giả thiết \(a^2+b^2+c^2+d^2=1\Rightarrow0< a,b,c,d< 1\)
Ta có: \(2\left(1-a\right)\left(1-b\right)=2-2\left(a+b\right)+2ab=a^2+b^2+c^2+d^2+1\)\(-2a-2b+2ab-2cd+2cd=\left(a+b-1\right)^2+\left(c-d\right)^2+2cd\ge2cd\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\ge cd\)(*)
Tương tự ta có: \(\left(1-c\right)\left(1-d\right)\ge ab\)(**)
Nhân theo từng vế cùng chiều của hai BĐT (*) và (**), ta được: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)\ge abcd\)
Đẳng thức xảy ra khi \(a=b=c=d=\frac{1}{2}\)
Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Đặt a+b=x;b+c=y;c+a=z
\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)
Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)
Dean thật, gõ gần xong rồi tự nhiên nó tạch, phải gõ lại -.-
Từ gt, ta suy ra:
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right].\dfrac{1}{2}=0\)(Tự phân tích, không còn kiên nhẫn để gõ lại)
Mà a+b+c khác 0 => a=b=c
Thay vào thì C=8
bai 2 :
dat cac tich ab , bc , ca lan luot la x,y,z ( khac 0 )
thay vao ta dc : x^3+y^3+z^3=3xyz
=> (x+y)(x^2-2xy+y^2)+z^3-3xyz=0
=>(x+y)(x^2+2xy+y^2)+z^3-3xy(x+y)-3xyz=0
=》（x+y+z)【（x+y）^2 -(x+y)z+z^2】-3xy(x+y+z)=0
=>(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0
=>\(\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)=0
=> x+y+z=0 hoac x=y=z
TH1 : a+b+c=0
=>P=-1
TH2 : a=b=c
=>P=8
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
Ta có:
\(\sqrt{2012}=abc+bcd+cda+dab-a-b-c-d=\left(bc-1\right)\left(a+d\right)+\left(ad-1\right)\left(b+c\right)\)
\(\Leftrightarrow2012=\left[\left(bc-1\right)\left(a+d\right)+\left(ad-1\right)\left(b+c\right)\right]^2\)
\(\le\left[\left(bc-1\right)^2+\left(b+c\right)^2\right]\left[\left(ad-1\right)^2+\left(a+d\right)^2\right]\)
\(=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\)
\(GT\Leftrightarrow2012=\left[\left(bc-1\right)\left(a+d\right)+\left(a+c\right)\left(ad-1\right)\right]^2\le\left[\left(bc-1\right)^2+\left(b+c^2\right)\right]\)
\(\left[\left(ad-1\right)^2+\left(a+d\right)^2\right]=\left(b^2+1\right)\left(c^2+1\right)\left(a^2+1\right)\left(d^2+1\right)\)
P/s: Mình không chắc đâu ! Tham khảo nha!
a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)
Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)
Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)
\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)
\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)
\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)
\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)
Cộng các vế lại, ta được :
\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)
\(\Rightarrow B\le6\)
Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)