Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 4:
a) C/m tương đương
\(\dfrac{a+b}{2}\ge\sqrt{ab}\) \(\Leftrightarrow a+b-2\sqrt{ab}\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\ge0\) => luôn đúng
=> \(\dfrac{a+b}{2}\ge\sqrt{ab}\Rightarrowđpcm\)
b) \(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\ge a+b+c\)
Áp dụng BĐT: \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)
+) \(\dfrac{bc}{a}+\dfrac{ba}{c}=b\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2b\)
+) \(\dfrac{ca}{b}+\dfrac{cb}{a}=c\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2c\)
+) \(\dfrac{ab}{c}+\dfrac{ac}{b}=a\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2a\)
Cộng vế vs vế ta có:
\(2\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\ge a+b+c\Rightarrowđpcm\)
c) Áp dụng BĐT Cô-si cho 2 số không âm ta có:
\(12^2=\left(3a+5b\right)^2\ge4.3a.5b=60ab\)
=> \(ab\le\dfrac{12}{5}\)
Vậy GTLN của P là \(\dfrac{12}{5}\)
Dấu ''=" xảy ra khi \(3a=5b\), từ đó ta có hệ
\(\left\{{}\begin{matrix}3a=5b\\3a+5b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{6}{5}\end{matrix}\right.\)
b) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3a^2+3b^2+3c^2\)
\(\Leftrightarrow0\le3a^2-a^2+3b^2-b^2+3c^2-c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0\le2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0\le\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
=> Đúng
Chúc bạn học tốt !!
a ) \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)
\(\Leftrightarrow0\le2a^2-a^2+2b^2-b^2-2ab\)
\(\Leftrightarrow0\le a^2-2ab+b^2\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
\(\Rightarrow\) đúng
Câu 6:
a: \(\left(a+1\right)^2>=4a\)
\(\Leftrightarrow a^2+2a+1-4a>=0\)
\(\Leftrightarrow a^2-2a+1>=0\)
\(\Leftrightarrow\left(a-1\right)^2>=0\)(luôn đúng)
b: \(\left\{{}\begin{matrix}a+1\ge2\sqrt{a}\\b+1\ge2\sqrt{b}\\c+1\ge2\sqrt{c}\end{matrix}\right.\)(Theo BĐT COSI)
\(\Leftrightarrow\left(a+1\right)\left(b+2\right)\left(c+1\right)\ge8\sqrt{abc}=8\)
a)\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)
\(\Leftrightarrow0\le2a^2-a^2+2b^2-b^2-2ab\)
\(\Leftrightarrow0\le a^2-2ab+b^2\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
=> Đúng
b) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3a^2+3b^2+3c^2\)
\(\Leftrightarrow0\le3a^2-a^2+3b^2-b^2+3c^2-c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0\le2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0\le\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
=> Đúng
a,Ta có : \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
Do : \(\left(a-b\right)^2\ge0\)nên \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\).
b, Xét : \(\left(a+b+c\right)^2+\left(a-b\right)^2-\left(a-c\right)^2+\left(b-c\right)^2\) . Khai triển và rút gọn, ta được :
\(3\left(a^2+b^2+c^2\right)\) . Vậy : \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)