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a ) CM : \(a^4+b^4\ge a^3b+b^3a\)
Giả sử điều cần c/m là đúng
\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)
\(\Rightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Rightarrow\left(a^3-b^3\right)\left(a-b\right)\ge0\)
\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
Ta có : \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)
\(\Rightarrow a^4+b^4\ge a^3b+b^3a\)
\(\Rightarrow2\left(a^4+b^4\right)\ge a^4+a^3b+b^4+b^3a\)
\(\Rightarrow2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\left(đpcm\right)\)
b ) \(\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(=a^4+a^3b+a^3c+b^3a+b^4+b^3c+c^3a+c^3b+c^4\)
\(=\left(a^4+b^4+c^4\right)+\left(a^3b+b^3a\right)+\left(b^3c+c^3b\right)+\left(a^3c+c^3a\right)\)
CMTT như a ) : \(\left\{{}\begin{matrix}a^4+b^4\ge a^3b+b^3a\\b^4+c^4\ge b^3c+c^3b\\a^4+c^4\ge a^3c+c^3a\end{matrix}\right.\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)\ge a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge a^4+b^4+c^4+a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\left(đpcm\right)\)
1. BĐT tương đương với \(6\left(a^2+b^2\right)-2ab+8-4\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge0\)
\(\Leftrightarrow\left[a^2-4a\sqrt{b^2+1}+4\left(b^2+1\right)\right]+\left[b^2-4b\sqrt{a^2+1}+4\left(a^2+1\right)\right]\)\(+\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-2\sqrt{b^2+1}\right)^2+\left(b-2\sqrt{a^2+1}\right)^2+\left(a-b\right)^2\ge0\)(đúng)
=> Đẳng thức không xảy ra
2. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)
\(\Leftrightarrow a^4+b^4+c^2+1\ge2a^2b^2-2a^2+2ac+2a\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^2-2ac+a^2\right)+\left(a^2-2a+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c-a\right)^2+\left(a-1\right)^2\ge0\)
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
Lời giải:
Xét hiệu:
\(2(a^4+b^4)-(a+b)(a^3+b^3)=2(a^4+b^4)-(a^4+ab^3+a^3b+b^4)\)
\(=a^4+b^4-a^3b-ab^3=(a^4-a^3b)-(ab^3-b^4)\)
\(=a^3(a-b)-b^3(a-b)=(a^3-b^3)(a-b)=(a-b)(a^2+ab+b^2)(a-b)\)
\(=(a-b)^2(a^2+ab+b^2)\)
Vì : \((a-b)^2\geq 0, \forall a,b\in\mathbb{R}\)
\(a^2+ab+b^2=(a+\frac{b}{2})^2+\frac{3}{4}b^2\geq 0, \forall a,b\in\mathbb{R}\)
\(\Rightarrow 2(a^4+b^4)-(a+b)(a^3+b^3)=(a-b)^2(a^2+ab+b^2)\geq 0\)
\(\Rightarrow 2(a^4+b^4)\geq (a+b)(a^3+b^3)\)
Ta có đpcm.
1) a) \(\left(a-b\right)^2-\left(a+b\right)^2=\left(a-b-a-b\right)\left(a-b+a+b\right)\)
\(=-2b\left(2a\right)=-4ab\)
b) ta có : \(\left(a+2b\right)^2+\left(b-a\right)^2-\left(a-b\right)^2=\left(a+2b\right)^2+\left(b-a\right)-\left(b-a\right)^2\)
\(=\left(a+2b\right)^2\)
2) ta có : \(\left(a-b\right)^2=\left(-\left(b-a\right)\right)^2=\left(b-a\right)^2\left(đpcm\right)\)
3) \(\left(a-b\right)^4=\left(a-b\right)^2\left(a-b\right)^2=\left(a^2-2ab+b^2\right)\left(a^2-2ab+b^2\right)\)
\(=a^4-2a^3b+a^2b^2-2a^3b+4a^2b^2-2ab^3+b^2a^2-2ab^3+b^4\)
\(=a^4-4a^3b+6a^2b^2-4ab^3+b^4\)
1)
\(\left(a+2b\right)^2+\left(b-a\right)^2-\left(a-b\right)^2\)
\(=\left(a^2+2a.2b+\left(2b\right)^2\right)+\left(b^2-2ba+a^2\right)-\left(a^2-2ab+b^2\right)\)
\(=a^2+4ab+4b^2+b^2-2ab+a^2-a^2+2ab-b^2\)
\(=a^2+4ab+4b^2\)
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm