Lời giải:
Áp dụng BĐT Bunhiacopxky kết hợp Cauchy ngược dấu ta có:
\((a\sqrt{b-1}+b\sqrt{a-1})^2=(\sqrt{a}.\sqrt{ab-a}+\sqrt{b}.\sqrt{ba-b})^2\leq (a+b)(ab-a+ba-b)\)
\(\leq \left(\frac{a+b+ab-a+ba+b}{2}\right)^2=(ab)^2\)
\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\leq ab\)
Ta có đpcm
Dấu "=" xảy ra khi \(a=b=2\)
Bài 1 : Cho a>c , b>c ( a,b,c>0). Cmr : \(\sqrt{c\sqrt{a-c}}+\sqrt{c\sqrt{b-c}}\le\sqrt{ab}\) (Hướng dẫn : chia cả 2 vế cho \(\sqrt{ab}\) , dùng cô-si)
Bài 2 : Cho \(a\ge1;b\ge1\) . Cmr \(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
Bài 3 : Tìm GTNN của \(A=\left(a+1\right)^2+\left(\frac{a^2}{a+1}+2\right)^2\) với mọi a\(\ne1\)
a)Cho 0 < c ; c < b ; b < a . CMR:\(\sqrt{c\left(a-c\right)}+\sqrt{b\left(b-c\right)}\le\sqrt{ab}\)
b)Cho \(x\ge1;y\ge1\). CMR:\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
CMR:\(a\sqrt{\left(b-1\right)}+b\sqrt{\left(a-1\right)}\le ab\) \(\forall a,b\ge1\)
Cho a, b, c dương thỏa \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\). Cmr: \(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}\ge1\)
cho a,b,c \(\ge\)0 thỏa a+b+c=1.CMR
\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Cho a,b là 2 số thực dương. CMR: \(\sqrt{\left(1+a\right)\left(1+b\right)}\ge1+\sqrt{ab}\)
Cho a, b >1
CMR :
\(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
cho a,b,c>0 thỏa mãn a+b+c=1. CMR: \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{3}{2}\)
1. a) cho \(1\le a,b,c\le2\). Tìm max \(P=\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\)
b) \(\left\{{}\begin{matrix}a,b,c\ge0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\sqrt{\frac{3a^2+1}{3b^2+1}}+\sqrt{\frac{3b^2+1}{3c^2+1}}+\sqrt{\frac{3c^2+1}{3a^2+1}}\le\frac{7}{2}\)
2.a) \(a,b\ge0;c\ge1;a+b+c=2\). cmr: \(\left(6-a^2-b^2-c^2\right)\left(2-abc\right)\le8\)
b) \(\left\{{}\begin{matrix}a+b\le2\\a^2+b^2+ab=3\end{matrix}\right.\). Tìm max,min \(P=a^2+b^2-ab\)
