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\(\frac{27}{3\sqrt{3x-2}+6}+\frac{8+4x-x^2}{x\sqrt{6-x}+4}\ge\frac{3}{2}+\frac{2x-14}{3\sqrt{6-x}+2}>0\)
Nên phần còn lại vô nghiệm
1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
@Ace Legona: sir tra hộ e câu này đúng hay sai đề vs ,nhẩm mãi không ra điểm rơi
Áp dụng bđt AM-GM có:
\(1+\dfrac{y}{z}\ge2\sqrt{\dfrac{y}{z}};1+\dfrac{z}{x}\ge2\sqrt{\dfrac{z}{x}}\)
Dễ dàng suy ra: \(M\ge\dfrac{x}{y}+2\sqrt{2}\cdot\sqrt[4]{\dfrac{y}{z}}+3\sqrt[3]{2}\cdot\sqrt[6]{\dfrac{z}{x}}=\dfrac{1}{\sqrt{2}}\left(\dfrac{x}{y}+4\sqrt[4]{\dfrac{y}{z}}+6\sqrt[6]{\dfrac{z}{x}}\right)+\left(1-\dfrac{1}{\sqrt{2}}\right)\cdot\dfrac{x}{y}+\left(3\sqrt[3]{2}-3\sqrt{2}\right)\cdot\sqrt[6]{\dfrac{z}{x}}\)
Theo AM-GM có: \(\dfrac{1}{\sqrt{2}}\left(\dfrac{x}{y}+4\sqrt[4]{\dfrac{y}{z}}+6\sqrt[6]{\dfrac{z}{x}}\right)\ge\dfrac{1}{2}\cdot11\sqrt[11]{\dfrac{x}{y}\cdot\dfrac{y}{z}\cdot\dfrac{z}{x}}=\dfrac{11}{\sqrt{2}}\) (1)
Theo đề: \(x\ge max\left\{y,z\right\}\) ta có: \(\left\{{}\begin{matrix}\dfrac{x}{y}\ge1\\\dfrac{z}{x}\le1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(1-\dfrac{1}{\sqrt{2}}\right)\cdot\dfrac{x}{y}\ge1-\dfrac{1}{\sqrt{2}}\left(2\right)\\\left(3\sqrt[3]{2}-3\sqrt{2}\right)\cdot\sqrt[6]{\dfrac{z}{x}}\ge3\sqrt[3]{2}-3\sqrt{2}\left(3\right)\end{matrix}\right.\)
Cộng theo vế bđt (1), (2) ,(3) có:\(A\ge\dfrac{11}{\sqrt{2}}+1-\dfrac{1}{\sqrt{2}}+3\sqrt[3]{2}-3\sqrt{2}=1+2\sqrt{2}+3\sqrt[3]{2}\)
Xảy ra khi \(x=y=z\)
Lâu lâu k đi khủng bố tinh thần :3
Ta đi cm \(1+2\sqrt{2}+3\sqrt[3]{2}\) là Min nhé
\(M'(x)=\dfrac{1}{y}+\dfrac{-\dfrac{z}{x^2}}{\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}}=\dfrac{x^2\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}-yz}{y\sqrt[3]{\left(1+\dfrac{z}{x}\right)^2}}\ge0\)
Vì vậy ta cần xét 2 trường hợp
*)\(y\ge z;x=y\). Đặt \(\dfrac{y}{z}=t\). Khi đó \(t\ge 1\) và cần cm \(f(t)\ge 0\)
\(f(t)=2\sqrt{1+t}+3\sqrt[3]{1+\dfrac{1}{t}}-2\sqrt{2}-3\sqrt[3]{2}\)
Thật vậy \(f'(t)=\dfrac{1}{\sqrt{1+t}}+\dfrac{-\dfrac{1}{t^2}}{\sqrt[3]{1+\dfrac{1}{t}}}=\dfrac{\sqrt[3]{t^4(t+1)^2}-\sqrt{1+t}}{\sqrt{1+t}\sqrt[3]{t^4(t+1)^2}}>0\)
\(\Rightarrow f(t)\ge f(1)=0\)
*)\(z\ge y ;x=z\). Khi đó \(t\ge 1\) và ta cm \(g(t)\ge 0\)
\(g(t)=t+2\sqrt{1+\dfrac{1}{t}}-1-2\sqrt{2}\)
Và \(g'(t)=1+\dfrac{-\dfrac{1}{t^2}}{\sqrt{1+\dfrac{1}{t}}}=\dfrac{\sqrt{t^3(t+1)}-1}{\sqrt{t^3(t+1)}}>0\)
Tức là \(g(t)\geq g(1)=0\)
Lời giải:
Vì $xy+yz+xz=1$ nên:
\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
\(y^2+1=y^2+xy+yz+xz=(y+x)(y+z)\)
\(z^2+1=z^2+xy+yz+xz=(z+y)(z+x)\)
Do đó:
\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)
\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{\sqrt{(x+y)^2(y+z)^2(x+z)^2}}\)
\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)
Lời giải:
Vì $xy+yz+xz=1$ nên:
\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
\(y^2+1=y^2+xy+yz+xz=(y+x)(y+z)\)
\(z^2+1=z^2+xy+yz+xz=(z+y)(z+x)\)
Do đó:
\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)
\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{\sqrt{(x+y)^2(y+z)^2(x+z)^2}}\)
\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)
Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
thật vĩ đại Hung nguyen
\(A=\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\)
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