Cho x,y,z là các số thực dương, chứng minh \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
Cho x,y,z là các số dương. Chứng minh rằng:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\) \(\ge\) \(\dfrac{2}{\sqrt{xy}}\) (1)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\) (2)
\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\) (3)
Cộng (1);(2);(3) vế theo vế ta được:
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\) (đpcm)
Cho các số thực dương $x,y,z$ thỏa mãn $x+y+z=1$. Chứng minh rằng:
\(\dfrac{x}{x+\sqrt{x+yz}}+\dfrac{y}{y+\sqrt{y+xz}}+\dfrac{z}{z+\sqrt{z+xy}}\le1\)
\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
Cho 3 số x, y, z dương TM: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\). CMR:
\(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Lời giải:
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+xz=xyz\)
\(\Rightarrow x^2+xy+yz+xz=x^2+xyz=x(x+yz)\)
\(\Leftrightarrow x+yz=\frac{x^2+xy+yz+xz}{x}=\frac{(x+y)(x+z)}{x}\)
\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\)
Áp dụng BĐT Bunhiacopxky:\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)
\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}\)
Hoàn toàn tương tự:
\(\sqrt{y+xz}\geq \frac{y+\sqrt{xz}}{\sqrt{y}}\); \(\sqrt{z+xy}\geq \frac{z+\sqrt{xy}}{\sqrt{z}}\)
Cộng theo vế các BĐT đã thu được ta có:
\(\text{VT}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}+\frac{y+\sqrt{xz}}{\sqrt{y}}+\frac{z+\sqrt{xy}}{\sqrt{z}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)
\(\Leftrightarrow \text{VT}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xyz}{\sqrt{xyz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}=\text{VP}\)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(x=y=z=3\)
cho x,y,z ≥ 0, chứng minh
1)\(\dfrac{1}{\sqrt{x+y}}\ge\dfrac{4}{4+x+y}\)
2)\(\dfrac{1}{xy}+\dfrac{1}{xz}\ge\dfrac{4}{x^2+yz}\)
Chứng minh bằng phép biến đổi tương đương:
1.
\(\Leftrightarrow4+x+y\ge4\sqrt{x+y}\)
\(\Leftrightarrow x+y-4\sqrt{x+y}+4\ge0\)
\(\Leftrightarrow\left(\sqrt{x+y}-2\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng
2.
\(\Leftrightarrow\dfrac{y+z}{xyz}\ge\dfrac{4}{x^2+yz}\)
\(\Leftrightarrow\left(y+z\right)\left(x^2+yz\right)\ge4xyz\)
\(\Leftrightarrow x^2y+x^2z+y^2z+z^2y-4xyz\ge0\)
\(\Leftrightarrow y\left(x^2+z^2-2xz\right)+z\left(x^2+y^2-2xy\right)\ge0\)
\(\Leftrightarrow y\left(x-z\right)^2+z\left(x-y\right)^2\ge0\) (đúng)
cho các số thực dương x,y,z thoả mãn \(\sqrt{x}\) + \(\sqrt{y}\) + \(\sqrt{z}\) = 1
chứng minh rằng : \(\sqrt{\dfrac{xy}{x+y+2z}}\) + \(\sqrt{\dfrac{yz}{y+z+2x}}\) + \(\sqrt{\dfrac{zx}{z+x+2y}}\) ≤ \(\dfrac{1}{2}\)
Cho x ≥ 1; y ≥ 2; z ≥ 3 và \(M=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
Chứng minh M ≤ \(\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\)
\(M=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{yz\sqrt{x-1}}{xyz}+\dfrac{xz\sqrt{y-2}}{xyz}+\dfrac{xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\)\(\Rightarrow\dfrac{\sqrt{x-1}}{x}\le\dfrac{x}{2}\cdot\dfrac{1}{x}=\dfrac{1}{2}\)
\(\sqrt{y-2}=\dfrac{\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{y}{2\sqrt{2}}\)\(\Rightarrow\dfrac{\sqrt{y-2}}{y}\le\dfrac{y}{2\sqrt{2}}\cdot\dfrac{1}{y}=\dfrac{1}{2\sqrt{2}}\)
\(\sqrt{z-3}=\dfrac{\sqrt{3\left(z-3\right)}}{\sqrt{3}}\le\dfrac{z}{2\sqrt{3}}\)\(\Rightarrow\dfrac{\sqrt{z-3}}{z}\le\dfrac{z}{2\sqrt{3}}\cdot\dfrac{1}{z}=\dfrac{1}{2\sqrt{3}}\)
Cộng theo vế 3 BĐT trên ta có:
\(M\le\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\) (ĐPCM)
cho các số thực dương thoả mãn: \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\)
CMR: \(\sqrt{\dfrac{xy}{x+y+2z}}+\sqrt{\dfrac{yz}{y+z+2x}}\sqrt{\dfrac{zx}{z+x+zy}}\le\dfrac{1}{2}\)
Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)
Tương tự cũng có:
\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)
\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
Cộng vế với vế ta được:
\(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)
cho x,y,z >0 tính:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
áp dụng BĐT cô si cho 2 số ko âm
\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}=\dfrac{2}{\sqrt{xy}}\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\)
cộng các vế vs nhau ta đc
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\) (đpcm)
Cho các số thực dương x, y, z thỏa mãn : xyz=1.CMR:
\(\dfrac{1}{\left(\sqrt{xy}+\sqrt{x}+1\right)^2}+\dfrac{1}{\left(\sqrt{yz}+\sqrt{y}+1\right)^2}+\dfrac{1}{\left(\sqrt{xz}+\sqrt{z}+1\right)^2}\ge\dfrac{1}{3}\)
Giúp mk với , mk sắp thi r...
Lời giải:
Đặt \((\sqrt{x}, \sqrt{y}, \sqrt{z})=(a,b,c)\Rightarrow abc=1\)
Bài toán trở thành chứng minh:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}\)
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Áp dụng 1 kết quả quen thuộc của BĐT AM-GM: \(x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\) ta có:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)^2\)
Mà:
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=\frac{c}{abc+ac+c}+\frac{ac}{bc.ac+b.ac+ac}+\frac{1}{ac+c+1}\)
\(=\frac{c}{1+ac+c}+\frac{ac}{c+1+ac}+\frac{1}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1\) (thay $abc=1$)
Do đó:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}.1^2=\frac{1}{3}\) (đpcm)
Dâu bằng xảy ra khi $a=b=c=1$ hay $x=y=z=1$