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\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
Tương tự \(\dfrac{\sqrt{1+y^3+z^3}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}};\dfrac{\sqrt{1+x^3+z^3}}{xz}\ge\dfrac{\sqrt{3}}{\sqrt{xz}}\)
\(\Rightarrow VT\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.\dfrac{3}{\sqrt[3]{xyz}}=3\sqrt{3}\)
Dấu "=" xảy ra khi x=y=z=1
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 \(x=y=z=3\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sqrt{\frac{xy}{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{z+y}\right)\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(\sum \sqrt{\frac{xy}{xy+z}}\leq \frac{1}{2}\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}\right)=\frac{3}{2}\)
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
\(A=\Sigma\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\Sigma\dfrac{\sqrt{3\sqrt[3]{1.x^3.y^3}}}{xy}\) (bđt Cô-si cho 3 số)
=> \(A\ge\Sigma\dfrac{\sqrt{3xy}}{xy}=\Sigma\dfrac{\sqrt{3}}{\sqrt{xy}}\ge3\sqrt[3]{\dfrac{\sqrt{3}}{\sqrt{xy}}.\dfrac{\sqrt{3}}{\sqrt{yz}}.\dfrac{\sqrt{3}}{\sqrt{zx}}}=3\sqrt{3}\) (bđt Cô-si cho 3 số)
Dấu "=" xảy ra <=> x = y = z = 1
Đặt
\(\left\{{}\begin{matrix}\sqrt{x}=a\\2\sqrt{y}=b\\3\sqrt{z}=c\end{matrix}\right.\)
\(\Rightarrow\frac{2}{a+b+c}-\frac{1}{ab+bc+ca}=\frac{1}{3}\)
\(\left(\sum a,\sum ab\right)\rightarrow\left(p,q\right)\)
Ta chứng minh :
\(\frac{2}{p}-\frac{1}{q}\le\frac{1}{3}\)
\(\Leftrightarrow p\ge\frac{6q}{q+3}\Leftrightarrow p^2\ge\frac{36q^2}{\left(q+3\right)^2}\)
Thấy : \(p^2\ge3q\)
Ta chứng minh :
\(3q\ge\frac{36q^2}{\left(q+3\right)^2}\Leftrightarrow\left(q-3\right)^2\ge0\)(luôn đúng).
\(\Rightarrow\)Dấu "=" xảy ra \(\Rightarrow a=b=c=1\)
\(\Rightarrow\left(x,y,z\right)\rightarrow\left(..,..,..\right)\)
#Kaito#
Lời giải:
Đặt \((\frac{1}{x}; \frac{1}{y}; \frac{1}{z})=(a,b,c)\). Bài toán trở thành:
Cho $a,b,c>0$ thỏa mãn $a+b+c=1$. CMR:
\(\frac{\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}}{\sqrt{abc}}\geq \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}(*)\)
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Do $a+b+c=1$ nên ta có:
\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}=\sqrt{a(a+b+c)+bc}+\sqrt{b(a+b+c)}+\sqrt{c(a+b+c)+ab}\)
\(=\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}\)
Mà áp dụng BĐT Bunhiacopxky:
\(\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\geq \sqrt{(a+\sqrt{bc})^2}+\sqrt{(b+\sqrt{ac})^2}+\sqrt{(c+\sqrt{ab})^2}\)
\(=a+\sqrt{bc}+b+\sqrt{ac}+c+\sqrt{ab}=a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
Vậy:\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\geq 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow \frac{\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}}{\sqrt{abc}}\geq \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)
$(*)$ được cm. BĐT hoàn thành. Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$ hay $x=y=z=3$
Áp dụng BĐT AM-GM, Ta có
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\Rightarrow yz\sqrt{x-1}\le\dfrac{xyz}{2}\)
Mà \(xz\sqrt{y-2}\le\dfrac{xz\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\)
\(yx\sqrt{z-3}\le yx.\dfrac{3+z-3}{2\sqrt{3}}=\dfrac{xyz}{2\sqrt{3}}\)
\(\Rightarrow\dfrac{xy\sqrt{x-1}+xz\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)
Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được
\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)
Ta phải chứng minh:
\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)
\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)
Theo C.B.S
\(\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Phải chứng minh
\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)
Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)
\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
=> ĐPCM
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+zx=xyz\)
\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Bình phương vế trái :
\(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\)
\(=\left(x+y+z+xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)Bình phương vế phải :
\(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=\left(xyz+x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
Suy ra cần phải chứng minh : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)(*)
Thật vậy, theo bđt Bunhiacopxki ta có : \(\sqrt{x+yz}.\sqrt{y+zx}\ge\sqrt{xy}+z\sqrt{xy}\)
\(\sqrt{y+zx}.\sqrt{z+xy}\ge\sqrt{yz}+x\sqrt{yz}\)
\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)
Cộng các bđt trên theo vế ta chứng minh được (*) đúng.
Vậy bđt ban đầu được chứng minh.
Ý tưởng khác
Cũng từ giả thiết suy ra \(xyz=xy+yz+xz\)
Suy ra \(\sqrt{x+yz}=\sqrt{\frac{x^2+xyz}{x}}=\sqrt{\frac{x^2+xy+yz+xz}{x}}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\)
Theo BĐT Cauchy-Schwarz ta có \(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\) do đó:
\(\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{x}=\sqrt{x}+\sqrt{\frac{yz}{x}}\)
Tương tự cho 2 BĐT còn lại \(\sqrt{y+xz}\ge\sqrt{y}+\sqrt{\frac{xz}{y}};\sqrt{z+xy}\ge\sqrt{z}+\sqrt{\frac{xy}{z}}\)
Cộng theo vế 3 BĐT được \(VT\ge\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)
\(\Leftrightarrow VT\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)
\(\Leftrightarrow VT\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\) (Đpcm)
P = 1
Sqrt(10P - 1) = sqrt(10.1-1)=3
\(P=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{3\sqrt{z}}{\sqrt{zx}+3\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}\sqrt{z}}{\sqrt{zx}+\sqrt{xyz}\sqrt{z}+\sqrt{xyz}}\)
\(=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)
\(=\dfrac{1+\sqrt{y}+\sqrt{yz}}{1+\sqrt{y}+\sqrt{yz}}=1\)
\(\Rightarrow\sqrt{10P-1}=\sqrt{10.1-1}=\sqrt{9}=3\)