Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(b=xa;c=ya\Rightarrow a^2+2x^2a^2\le3y^2a^2\Leftrightarrow1+2x^2\le3y^2\)
Ta cần chứng minh:\(\frac{1}{a}+\frac{2}{xa}\ge\frac{3}{ya}\Leftrightarrow1+\frac{2}{x}\ge\frac{3}{y}\)
Vậy ta viết được bài toán thành dạng đơn giản hơn:
Cho x, y > 0 thỏa mãn \(1+2x^2\le3y^2\). Chứng minh:\(1+\frac{2}{x}\ge\frac{3}{y}\)
Tối về em suy nghĩ tiếp ạ!
Áp dụng BĐT bu-nhi-a ta có \(\left(a+2b\right)^2\le3\left(a^2+2b^2\right)\le9c^2\Rightarrow a+2b\le3c\)
=>\(\frac{1}{a+2b}\ge\frac{1}{3c}\Rightarrow\frac{9}{a+2b}\ge\frac{3}{c}\)
Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{3}{c}\Rightarrow\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\left(ĐPCM\right)\)
8n
1,
\(A=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)
\(\ge1+1+2\sqrt{\frac{a}{b}.\frac{b}{a}}+a+b+\frac{a+b}{ab}=4+a+b+\frac{4\left(a+b\right)}{\left(a+b\right)^2}=4+a+b+\frac{4}{a+b}\)
lại có \(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a+b\le\sqrt{2}\)
\(4+a+b+\frac{4}{a+b}=4+\left(a+b+\frac{2}{a+b}\right)+\frac{2}{a+b}\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
\(\Rightarrow A\ge4+3\sqrt{2}\)
câu 2
ta có:\(\left(2b^2+a^2\right)\left(2+1\right)\ge\left(2b+a\right)^2\Rightarrow3c\ge a+2b\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\left(Q.E.D\right)\)
Thì bạn cứ biết là áp dụng bđt
\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{\left(1+2\right)^2}{a+2b}=\frac{9}{a+2b}\) ( BĐT Schwarz )
Ta cần cm \(a+2b\le3c\)
\(\left(a+2b\right)^2=\left(1\cdot a+\sqrt{2}\cdot b\cdot\sqrt{2}\right)^2\le\left(1^2+\left(\sqrt{2}\right)^2\right)\left(a^2+2b^2\right)=3\left(a^2+2b^2\right)\le3.3c^2=9c^2\)( BUN nhiacopxki )
<=> \(\sqrt{\left(a+2b\right)^2}\le\sqrt{9c^2}\Leftrightarrow a+2b\le3c\) ( XONG )
Dấu '' = '' xảy ra khi a = b = c
Ta có: \(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{\left(1+1+1\right)^2}{a+b+b}=\frac{9}{a+2b}\)
Theo BĐT Bu-nhi-a-cốp-xki ta có:
\(\left(a+2b\right)^2=\left(1.a+\sqrt{2}.\sqrt{2}b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\le3.3c^2=9c^2\Rightarrow a+2b\le3c\)
\(\Rightarrow\frac{1}{a}+\frac{2}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
\(a+2b=1.a+\sqrt{2}.\sqrt{2}b\le\sqrt{\left(1+2\right)\left(a^2+2b^2\right)}\le\sqrt{3.3c^2}=3c\)
\(\Rightarrow a+2b\le3c\)
\(\Rightarrow\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\) (đpcm)
Dấu "=" khi \(a=b=c\)
Áp dụng bất đẳng thức cauchy- schawarz
\(\left(a^2+2b^2\right)3\ge\left(a+2b\right)^2\)
\(\Rightarrow a^2+2b^2\ge\frac{\left(a+2b\right)^2}{3}\)\(\Rightarrow\frac{\left(a+2b\right)^2}{3}\le3c^2\Leftrightarrow\left(a+2b\right)^2\le9c^2\Leftrightarrow a+2b\le3c\)
áp dụng bất đẳng thức Cauchy - schawarz dạng engel
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{\left(1+2\right)^2}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\overset{BĐT\text{ }Cô-si}{\ge}\frac{9}{a+b+b}=\frac{9}{a+2b}\)
Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki ta có:\(\left(a+2b\right)^2\le\left(1^2+\sqrt{2}^2\right)\left[a^2+\left(\sqrt{2}b\right)^2\right]=3\left(a^2+2b^2\right)\le9c^2\\ \Rightarrow a+2b\le3c\\ \Rightarrow\frac{1}{a}+\frac{2}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}a=b\\a^2+2b^2=3c^2\\\frac{a}{1}=\frac{\sqrt{2}b}{\sqrt{2}}\end{matrix}\right.\Rightarrow a=b=c\)