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\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)
\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)
\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)
=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{3\sqrt{z}}{\sqrt{zx}+3\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}\sqrt{z}}{\sqrt{zx}+\sqrt{xyz}\sqrt{z}+\sqrt{xyz}}\)
\(=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)
\(=\dfrac{1+\sqrt{y}+\sqrt{yz}}{1+\sqrt{y}+\sqrt{yz}}=1\)
\(\Rightarrow\sqrt{10P-1}=\sqrt{10.1-1}=\sqrt{9}=3\)
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 $x=y=z=3$
\(A=\Sigma\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\Sigma\dfrac{\sqrt{3\sqrt[3]{1.x^3.y^3}}}{xy}\) (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
\(xy+xz+yz=6xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y}=b\\\frac{1}{z}=c\end{matrix}\right.\) \(\Rightarrow a+b+c=6\)
\(T=\sum x\sqrt{\frac{x}{1+x^3}}=\sum\sqrt{\frac{x^3}{1+x^3}}=\sum\sqrt{\frac{1}{1+\frac{1}{x^3}}}=\sum\frac{1}{\sqrt{1+a^3}}=\sum\frac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\Rightarrow T\ge\sum\frac{2}{a+1+a^2-a+1}=\sum\frac{2}{a^2+2}\)
Ta có đánh giá: \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\) với mọi \(0< a< 6\)
Thật vậy, \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\Leftrightarrow18-\left(a^2+2\right)\left(7-2a\right)\ge0\)
\(\Leftrightarrow2a^3-7a^2+4a+4\ge0\)
\(\Leftrightarrow\left(a-2\right)^2\left(2a+1\right)\ge0\) luôn đúng với mọi \(0< a< 6\)
Tương tự ta có: \(\frac{2}{b^2+2}\ge\frac{7-2b}{9}\) ; \(\frac{2}{c^2+2}\ge\frac{7-2c}{9}\)
\(\Rightarrow T\ge\frac{21-2\left(a+b+c\right)}{9}=\frac{21-12}{9}=1\)
\(\Rightarrow T_{min}=1\) khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)
Với a,b,c dưog thì \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}>=\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
\(P>=\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+\sqrt{1+x^3}+\sqrt{1+y^3}+\sqrt{1+z^3}}\)
\(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+x^2\right)}< =\dfrac{2+x^2}{2}\)
Dấu = xảy ra khi x=2
=>\(P>=\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2+6}=\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2+6}\)
Đặt t=(x+y+z)^2(t>=36)
=>P>=2t/t-6
Xét hàm số \(f\left(t\right)=\dfrac{t}{t+6}\left(t>=36\right)\)
\(f'\left(t\right)=\dfrac{6}{\left(t+6\right)^2}>=0,\forall t>=36\)
=>f(t) đồng biến
=>f(t)>=f(36)=6/7
=>P>=12/7
Dấu = xảy ra khi x=y=z=2
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 \(xz\sqrt{y-2}\le\dfrac{xz\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\)
\(yx\sqrt{z-3}\le yx.\dfrac{3+z-3}{2\sqrt{3}}=\dfrac{xyz}{2\sqrt{3}}\)
\(\Rightarrow\dfrac{xy\sqrt{x-1}+xz\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)
nhờ mn giúp mk bài này vs ạ
mk đang cần gấp !
cảm ơn mn nhiều
Đặt \(\left(\sqrt[3]{x};\sqrt[3]{y};\sqrt[3]{z}\right)=\left(a;b;c\right)\) \(\Rightarrow a^6+b^6+c^6=3\)
\(a^6+a^6+a^6+a^6+a^6+1\ge6a^5\)
Tương tự: \(5b^6+1\ge6b^5\) ; \(5c^6+1\ge6c^5\)
Cộng vế với vế: \(18=5\left(a^6+b^6+c^6\right)+3\ge6\left(a^5+b^5+c^5\right)\)
\(\Rightarrow3\ge a^5+b^6+b^5\)
BĐT cần chứng minh: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge a^3b^3+b^3c^3+c^3a^3\)
Ta có:
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge a+b+c\) (1)
Mà \(3\left(a+b+c\right)\ge\left(a^5+b^5+c^5\right)\left(a+b+c\right)\ge\left(a^3+b^3+c^3\right)^2\ge3\left(a^3b^3+b^3c^3+c^3a^3\right)\)
\(\Rightarrow a+b+c\ge a^3b^3+b^3c^3+c^3a^3\) (2)
Từ (1);(2) \(\Rightarrow\) đpcm
\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
áp dụng BĐT AM-GM
\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)
\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)
có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)
\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)
\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)
tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)
(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
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