Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(T=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{\frac{4a+3a+b}{2}+\frac{4b+3b+a}{2}}=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}\)
\(\Rightarrow T_{min}=\frac{1}{2}\) khi \(a=b\)
\(\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\) \(\ge\frac{2\left(a+b\right)}{\frac{4a+3a+b}{2}+\frac{4b+3b+a}{2}}=\frac{2\left(a+b\right)}{\frac{8\left(a+b\right)}{2}}=\frac{1}{2}\)
dau = xay ra khi a=b
Bạn tham khảo:
Câu hỏi của Nguyễn Thị Hồng Ngọc - Toán lớp 9 | Học trực tuyến
+ \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)
Dấu "=" \(\Leftrightarrow4a=3a+b\Leftrightarrow a=b\)
+ \(\sqrt{4b\left(3b+a\right)}\le\frac{a+7b}{2}\) Dấu "=" \(\Leftrightarrow a=b\)
\(\Rightarrow\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}\le4\left(a+b\right)\)
\(\Rightarrow\frac{1}{2}Q=\frac{a+b}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{a+b}{4\left(a+b\right)}=\frac{1}{4}\)
\(\Rightarrow Q\ge\frac{1}{2}\)
Dấu "=" \(\Leftrightarrow a=b\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\)
\(\le\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}\)
\(=\sqrt{4\left(a+b\right)^2}=2\left(a+b\right)\)
\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Áp dụng Cauchy-Schwarz ta có:
\(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}=\frac{1}{2}\)
Ta có:
\(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\)
\(\ge\frac{2\left(a+b\right)}{\frac{4a+3a+b}{2}+\frac{4b+3b+a}{2}}=\frac{2\left(a+b\right)}{4\left(a+b\right)}=\frac{1}{2}\)
Dấu = xảy ra khi \(a=b\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}=\sqrt{a}\sqrt{3a+b}+\sqrt{b}\sqrt{3b+a}\)
\(\le\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}=2\left(a+b\right)\)
\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Đẳng thức xảy ra khi \(a=b\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}=\sqrt{a}\cdot\sqrt{3a+b}+\sqrt{b}\cdot\sqrt{3b+a}\)
\(\le\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}=2\left(a+b\right)\)
\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Xảy ra khi \(a=b\)
Ta có \(T=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\)
=> \(T\ge\frac{4\left(a+b\right)}{4a+3a+b+4b+3b+a}=\frac{1}{2}\)( vì \(\sqrt{4a\left(3a+b\right)}\le\frac{3a+4a+b}{2}\)
Vậy MinP=1/2 khi a=b