Lời giải:
Áp dụng BĐT Cauchy ngược dấu:
\(\sqrt{ab}=\frac{1}{2}\sqrt{a.4b}\leq \frac{1}{2}.\frac{a+4b}{2}=\frac{a+4b}{4}\)
\(\sqrt[3]{abc}=\frac{1}{4}\sqrt[3]{a.4b.16c}\leq \frac{1}{4}.\frac{a+4b+16c}{3}=\frac{a+4b+16c}{12}\)
Do đó:
\(a+\sqrt{ab}+\sqrt[3]{abc}\leq a+\frac{a+4b}{4}+\frac{a+4b+16c}{12}=\frac{4}{3}(a+b+c)=\frac{4}{3}\)
Vậy \(P_{\max}=\frac{4}{3}\)
Cho a,b,c>0 và a+b+c=1.Tìm max của P=\(a+\sqrt{ab}+\sqrt[3]{abc}\)
Cho a,b,c >0 t/m a+b+c=abc-2. Tìm max
\(P=\sqrt{\dfrac{1}{a+1}}+\sqrt{\dfrac{1}{b+1}}+\sqrt{\dfrac{1}{c+1}}\)
a,b,c>0 thỏa mãn a+b+c=1. tìm max \(P=a+\sqrt{ab}+\sqrt[3]{abc}\)
Cho a,b,c >0 và a+b+c = abc. Tìm Max
\(S=\dfrac{a}{\sqrt{cb\left(1+a^2\right)}}+\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}+\dfrac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Cho a,b,c >0 tm a+b+c=1.Tìm max \(S=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
Cho a, b, c > 0 thỏa mãn a + \(\sqrt{ab}+\sqrt[3]{abc}=\dfrac{4}{3}\)
Tìm GTNN của A = a + b + c
Cho a,b,c>0 thỏa a+b+c=3. Tìm Max P \(\frac{2}{3+ab+bc+ca}+\frac{\sqrt{abc}}{6} +\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cho x,y,z>0 thỏa \(3x+y+z=x^2+y^2+z^2+2xy\) . Tìm Min P= \(\frac{20}{\sqrt{x+2}}+\frac{20}{\sqrt{y+2}}+x+y+z\)
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
Cho a,b,c>0 ;a+b+c=1
Tìm max A= \(\dfrac{ab}{\sqrt{c+ab}}\) +\(\dfrac{bc}{\sqrt{a+bc}}\)+\(\dfrac{ca}{\sqrt{b+ca}}\)
