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Từ a3 + b3 + c3 = 3abc
<=> (a + b)(a2 - ab + b2) + c3 - 3abc = 0
<=> (a + b)3 + c3 - 3ab(a + b) - 3abc = 0
<=> (a + b + c)(a2 + 2ab + b2 - ac - bc + c2) - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> \(\orbr{\begin{cases}a+b+c=0\left(loại\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
<=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> a = b = c
=> tam giác đó là tam giác đều
b) Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
CM đúng (tự cm tđ)
Ta có: \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}=9\)(vì x + y + z = 1)
Dấu "=" xảy ra <=> x = y = z = 1/3
a) Vì a, b, c là độ dài ba cạnh của một tam giác => a, b, c > 0
Ta có : a3 + b3 + c3 = 3abc
<=> a3 + b3 + c3 - 3abc = 0
<=> ( a + b )3 - 3ab( a + b ) + c3 - 3abc = 0
<=> [ ( a + b )3 + c3 ] - [ 3ab( a + b ) + 3abc ] = 0
<=> ( a + b + c )( a2 + b2 + c2 + 2ab - ac - bc ) - 3ab( a + b + c ) = 0
<=> ( a + b + c )( a2 + b2 + c2 - ab - ac - bc ) = 0
<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
Dễ thấy không thể xảy ra trường hợp a + b + c = 0 vì a, b, c > 0
Xét TH còn lại ta có :
a2 + b2 + c2 - ab - ac - bc = 0
<=> 2(a2 + b2 + c2 - ab - ac - bc) = 2.0
<=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
<=> ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( c2 - 2ac + a2 ) = 0
<=> ( a - b )2 + ( b - c )2 + ( c - a )2 = 0 (*)
Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(c-a\right)^2\end{cases}}\ge0\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
=> Tam giác đó là tam giác đều ( đpcm )
Áp Dụng BĐT svacxơ, ta có
\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}=9\left(ĐPCM\right)\)
^_^
Đặt a = \(x^2+2yz\); b = \(y^2+2xz\); c = \(z^2+2xy\)
\(\Rightarrow\)\(a,b,c>0\)và \(a+b+c=\left(x=y+z\right)^2=1\)
+) C/m : \(\left(a=b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Rightarrow\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\)
Hay \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\)
\(\Rightarrow\)ĐPCM
hên xui thôi -_-
Áp dụng BĐT Cauchy-schwarz dạng engel,ta có:
\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1+1\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=\frac{9}{\left(x+y+z\right)^2}=9\)
\(\Rightarrowđpcm\)
Ta có :
\(x+y+z=1\)
\(\Rightarrow\left(x+y+z\right)^2=1\)
Áp dụng BĐT Cauchy-schwar dưới dạng engel ta có :
\(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}\ge\dfrac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\dfrac{9}{1}=9\)
\(\text{Ta có : }x+y+z=1\\ \Rightarrow\left(x+y+z\right)^2=1\\ \Rightarrow x^2+y^2+z^2+2xy+2xz+2yz=1\\ \Rightarrow\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\\ =\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{x^2+2yz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{y^2+2xz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{z^2+2xy}\\ =\dfrac{x^2+2yz}{x^2+2yz}+\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}+\dfrac{y^2+2xz}{y^2+2xz}+\dfrac{z^2+2xy}{y^2+2xz}+\dfrac{x^2+2yz}{z^2+2xy}+\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{z^2+2xy}\\ =1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\)Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\\ \ge1+2+2+1+2+1\ge9\left(đpcm\right)\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}y^2+2xz=x^2+2yz\\z^2+2xy=x^2+2yz\\y^2+2xz=z^2+2xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2-2yz=x^2-2xz\\z^2-2yz=x^2-2xy\\y^2-2xy=z^2-2xz\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-2yx+z^2=x^2-2xz+z^2\\z^2-2yz+y^2=x^2-2xy+y^2\\y^2-2xy+x^2=z^2-2xz+x^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-z\right)^2=\left(x-z\right)^2\\\left(z-y\right)^2=\left(x-y\right)^2\\\left(y-x\right)^2=\left(z-x\right)^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-z=x-z\\z-y=x-y\\y-x=z-x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\z=x\\y=z\end{matrix}\right.\Leftrightarrow x=y=z\\\text{Mà } x+y+z=1\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ \Leftrightarrow x=y=z=\dfrac{1}{3}\)
Vậy \(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\ge9\) với \(x;y;z>0\) và \(x+y+z=1\)
đẳng thức xảy ra khi : \(x=y=z=\dfrac{1}{3}\)
\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=1\)
Dấu "=" xảy ra khi \(x=y=z\)
\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}=9\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y+z=1\\x=y=z\end{cases}\Leftrightarrow x=y=z=\frac{1}{3}}\)
ban chung minh gium mik bdt nha