Cho a, b, c >0. CMR: \(\dfrac{a+b+c}{3}\) - \(\sqrt[3]{abc}\) ≤ \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2}{3}\)
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Bài 1:
Áp dụng BĐT AM-GM ta có:
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$
$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$
Cộng theo vế và thu gọn:
$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$
Ta có đpcm.
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Bài 2:
$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$
$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$
$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$
Cộng theo vế và rút gọn thu được:
$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 12 2020
Hai bài giống hệt nhau về cách làm:
Cho a, b, c > 0 thoả mãn: \(a b c=\sqrt{a} \sqrt{b} \sqrt{c}=2\). Chứng minh rằng: \(\dfrac{\sqrt{a}}{a 1} \dfrac{\sqrt{... - Hoc24
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 6 2021
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
26 tháng 8 2021
`sqrta+sqrtb+sqrtc=2`
`<=>(sqrta+sqrtb+sqrtc)^2=4`
`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`
`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`
`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`
`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`
Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`
`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`
`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`
`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`
`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`
`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`
DH
26 tháng 8 2021
a ơi giả thiết là a+b+c=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)=2 nhé a
26 tháng 5 2018
Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)
Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)
\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)
\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)
Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)
\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)
\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*
9 tháng 6 2018
B1:a)ĐK: \(x\ne 0;4;9\)
b)\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1+1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-9-x+4+x^{\dfrac{1}{2}}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x^{\dfrac{1}{2}}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)\(=\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}\)
c)Vì \(x^{\dfrac{1}{2}}+1>0\forall x\) nên
\(P< 0< =>x-2x^{\dfrac{1}{2}}< 0\)
\(\Leftrightarrow x^{\dfrac{1}{2}}\left(x^{\dfrac{1}{2}}-2\right)< 0\)
\(\Leftrightarrow0< x< 4\)
Vậy 0<x<4 thì P<0
d)tA CÓ: \(\dfrac{1}{P}=\dfrac{x-2x^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}+1}=\dfrac{x-2x^{\dfrac{1}{2}}+1-1}{x^{\dfrac{1}{2}}+1}=\dfrac{\left(x^{\dfrac{1}{2}}-1\right)^2-1}{x^{\dfrac{1}{2}}+1}\ge-1\)
"=" khi x=1
B2:
a)\(A=x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)-5\)
\(=\left(x-y\right)^2-1+4\left(x-y\right)-4\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y+5\right)\left(x-y-1\right)\)
b)\(P=x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+2\left(x^2+x\right)+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\ge0\forall x\)
Vậy MinP=0
c)\(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)
\(=\left(x^2+x-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=\left(1-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=0\left(x^4+x^3+2x^2+x+3\right)+4=4\)
Vậy x^2+x=1 thì Q=4
B3:a)\(2xy+x+y=83\)
\(\Leftrightarrow x\left(2y+1\right)+\dfrac{1}{2}\left(2y+1\right)=\dfrac{167}{2}\)
\(\Leftrightarrow2x\left(2y+1\right)+1\left(2y+1\right)=167\)
\(\Leftrightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Mà \(Ư\left(167\right)=\left\{\pm1;\pm167\right\}\)
\(\Leftrightarrow\left(x;y\right)=\left(-84;-1\right);\left(-1;-84\right);\left(0;83\right);\left(83;0\right)\)
Vậy...
b)\(y^2+2xy-3x-2=0\)
\(\Leftrightarrow x^2+y^2+2xy-x^2-3x-2=0\)
\(\Leftrightarrow\left(x+y\right)^2=x^2+3x+2\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\)
Vì \(x;y\in Z\) nên VT là số chính phương VP là tích 2 số nguyên liên tiếp
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)
Vậy...
B5:\(B=\dfrac{x^2+x+1}{x^2-x+1}\)
\(\Leftrightarrow x^2\left(B-1\right)+x\left(-B-1\right)+\left(B-1\right)=0\)
\(\Delta=\left(-B-1\right)^2-4\left(B-1\right)\left(B-1\right)\)
\(=-\left(B-3\right)\left(3B-1\right)\)
pt có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(B-3\right)\left(3B-1\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}B-3\le0\\3B-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}B\le3\\B\ge\dfrac{1}{3}\end{matrix}\right.\)
Min B=1/3 khi x=-1; Max B=3 khi x=1
HN
11 tháng 4 2017
Ta có:
\(\left(\sqrt{\dfrac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\dfrac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\dfrac{c^3}{5c^2+\left(a+b\right)^2}}\right)^2\le\left(a+b+c\right)\left(\dfrac{a^2}{5a^2+\left(b+c\right)^2}+\dfrac{b^2}{5b^2+\left(c+a\right)^2}+\dfrac{c^2}{5c^2+\left(a+b\right)^2}\right)\left(1\right)\)
Giờ ta chứng minh:
\(P=\dfrac{a^2}{5a^2+\left(b+c\right)^2}+\dfrac{b^2}{5b^2+\left(c+a\right)^2}+\dfrac{c^2}{5c^2+\left(a+b\right)^2}\le\dfrac{1}{3}\)
Ta có:
\(\dfrac{a^2}{5a^2+\left(b+c\right)^2}\le\dfrac{a^2}{9}\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{2a^2+bc}+\dfrac{1}{2a^2+bc}\right)=\dfrac{1}{9}\left(\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{2a^2}{2a^2+bc}\right)=\dfrac{1}{9}+\dfrac{1}{9}\left(\dfrac{a^2}{a^2+b^2+c^2}-\dfrac{bc}{2a^2+bc}\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{b^2}{5b^2+\left(c+a\right)^2}\le\dfrac{1}{9}+\dfrac{1}{9}\left(\dfrac{b^2}{a^2+b^2+c^2}-\dfrac{ca}{2b^2+ca}\right)\\\dfrac{c^2}{5c^2+\left(a+b\right)^2}\le\dfrac{1}{9}+\dfrac{1}{9}\left(\dfrac{c^2}{a^2+b^2+c^2}-\dfrac{ab}{2c^2+ab}\right)\end{matrix}\right.\)
Cộng vế theo vế ta được
\(P\le\dfrac{4}{9}-\dfrac{1}{9}\left(\dfrac{bc}{2a^2+bc}+\dfrac{ca}{2b^2+ca}+\dfrac{ab}{2c^2+ab}\right)\)
\(\le\dfrac{4}{9}-\dfrac{1}{9}.\dfrac{\left(ab+bc+ca\right)^2}{bc\left(2a^2+bc\right)+ca\left(2b^2+ca\right)+ab\left(2c^2+ab\right)}=\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{1}{3}\left(2\right)\)
Từ (1) và (2) ta có
\(\sqrt{\dfrac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\dfrac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\dfrac{c^3}{5c^2+\left(a+b\right)^2}}^2\le\sqrt{\dfrac{a+b+c}{3}}\)
AH
Akai Haruma
Giáo viên
29 tháng 9 2017
Lời giải:
Áp dụng hệ quả của BĐT AM-GM:
\(\text{VT}^2=\left[\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\right]^2\geq 3\left(\frac{1}{ab(a+1)(b+1)}+\frac{1}{bc(b+1)(c+1)}+\frac{1}{ca(a+1)(c+1)}\right)\)
\(\Leftrightarrow \text{VT}^2\geq 3.\frac{a^2+b^2+c^2+a+b+c}{abc(a+1)(b+1)(c+1)}\geq 3.\frac{a+b+c+ab+bc+ac}{abc(a+1)(b+1)(c+1)}\)
\(\Leftrightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(a+1)(b+1)(c+1)}\) \((1)\)
Ta sẽ cm \((a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3\). Thật vậy:
Áp dụng BĐT AM-GM:
\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}\)
Cộng theo vế: \(\Rightarrow 3\geq \frac{3(\sqrt[3]{abc}+1)}{\sqrt[3]{(a+1)(b+1)(c+1)}}\)
\(\Rightarrow (a+1)(b+1)(c+1)\geq (\sqrt[3]{abc}+1)^3\) (2)
Từ \((1),(2)\Rightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(1+\sqrt[3]{abc})^3}=\frac{9}{\sqrt[3]{a^2b^2c^2}(1+\sqrt[3]{abc})^2}=\text{VP}^2\)
\(\Leftrightarrow \text{VT}\geq \text{VP}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
