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Áp dụng BĐT AM - GM ta có:
\(4\sqrt{ab}=2\sqrt{a.4b}\le a+4b\)
\(4\sqrt{bc}=2\sqrt{b.4c}\le b+4c\)
\(4\sqrt[3]{abc}=\sqrt[3]{a.4b.16c}\le\frac{a+4b+16c}{3}\)
Cộng theo vế 3 BĐT ta được:
\(8a+3b+4\left(\sqrt{ab}+\sqrt{bc}+\sqrt[3]{abc}\right)\le\frac{28}{3}\left(a+b+c\right)\)
\(\Rightarrow P\le\frac{28\left(a+b+c\right)}{3+3\left(a+b+c\right)^2}=\frac{14}{3}-\frac{14\left(a+b+c-1\right)^2}{3\left[\left(a+b+c\right)^2+1\right]}\le\frac{14}{3}\)
\(\Rightarrow Max_P=\frac{14}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow a+b+c=1\)và \(a=4b=16c\)
\(\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}\)
\(A=\frac{1}{6}\left(6-2x\right)\left(12-3y\right)\left(2x+3y\right)\)
\(A\le\frac{1}{6}\left(\frac{6-2x+12-3y+2x+3y}{3}\right)^3=36\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
\(A=\frac{\frac{ab}{\sqrt{2}}\sqrt{2\left(c-2\right)}+\frac{bc}{\sqrt{3}}\sqrt{3\left(a-3\right)}+\frac{ca}{2}\sqrt{4\left(b-4\right)}}{abc}\)
\(A\le\frac{\frac{abc}{2\sqrt{2}}+\frac{abc}{2\sqrt{3}}+\frac{abc}{4}}{abc}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)
\(\frac{a^2}{\sqrt{3a^2+8b^2+12ab+2ab}}\ge\frac{a^2}{\sqrt{3a^2+9b^2+12ab+a^2+b^2}}=\frac{a^2}{\sqrt{\left(2a+3b\right)^2}}=\frac{a^2}{2a+3b}\)
\(\Rightarrow VT\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{1}{5}\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\)\(\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)\(\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)=\(\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(1)
Tương tự ta có: \(\frac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(2)
\(\frac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(3)
Cộng theo vế của (1);(2)&(3) ta đc:
A\(\le1\)
Dấu''='' xảy ra\(\Leftrightarrow\)a=b=c
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$