Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Vì $1=a^2+b^2+c^2$ nên:
\(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}=\frac{a^2+b^2+c^2}{a^2+b^2}+\frac{a^2+b^2+c^2}{b^2+c^2}+\frac{a^2+b^2+c^2}{c^2+a^2}\)
\(=3+\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\)
\(\leq 3+\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\) (theo BĐT AM-GM)
\(=3+\frac{a^3+b^3+c^3}{2abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{\frac{1}{3}}\)
ta có
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\) (vì a+b=c=abc)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow M=2\)
bài này xài karamata là đẹp nhất nè nhanh gọn lẹ mà ko bt bn học chưa
Ahaha :D giỡn xíu lớp 8 có khi AM-HM còn chưa học :3, bài này với bn phải xài khai triển Abel 
\(Q=\frac{1}{c+1}+\frac{ab+abc-c-1}{\left ( a+1 \right )\left ( b+1 \right )\left ( c+1 \right )}=\frac{1}{c+1}+\frac{ab-1}{\left ( a+1 \right )\left ( b+1 \right )}\)
\(=\frac{1}{c+1}+\frac{a}{a+1}+\frac{b}{b+1}-1=\frac{a}{a+1}+\frac{b}{b+1}-\frac{c}{c+1}\)
Dự đoán dấu "=" rơi khi \(a=b-1=c-2=1\) nên c/m
\(\frac{a}{a+1}+\frac{b}{b+1}-\frac{c}{c+1}\geq \frac{5}{12}\)
\(\Leftrightarrow \left ( \frac{a}{a+1}-\frac{1}{2} \right )+\left ( \frac{b}{b+1}-\frac{2}{3} \right )+\left ( \frac{3}{4}-\frac{c}{c+1} \right )\geq 0\)
\(\Leftrightarrow \frac{a-1}{2a+2}+\frac{b-2}{3b+3}+\frac{3-c}{4c+4}\geq 0\)
\(\Leftrightarrow \left ( 3-c \right )\left ( \frac{1}{4c+4}-\frac{1}{3b+3} \right )+\left ( 3-c+b-2 \right )\left ( \frac{1}{3b+3}-\frac{1}{2a+2} \right )+\left ( 3-c+b-2+a-1 \right )\frac{1}{2a+2}\geq 0\)
\(\Leftrightarrow \frac{\left ( c-3 \right )\left ( 4c-3b+1 \right )}{12\left ( b+1 \right )\left ( c+1 \right )}+\frac{\left ( b+1-c \right )\left ( 2a-3b-1 \right )}{6\left ( b+1 \right )\left ( a+1 \right )}+\frac{a+b-c}{2a+2}\geq 0\)
Hơi xấu nhỉ nhưng xong rồi đó :)
1/(a+b) + 1/(b+c) + 1/(c+a) = 4/(a+b+c)
=> [1/(a+b) + 1/(b+c) + 1/(c+a)](a+b+c) = 4
=> 3 + c/(a+b) +a/(b+c) + b/(c+a) = 4
=> [3 + c/(a+b) + a/(b+c) + b/(c+a)](a+b+c) = 4(a+b+c)
=> 3(a+b+c) + c + c2(a+b) + a + a2(b+c) + b + b2(c+a) = 4(a+b+c)
=> a2(b+c) + b2(c+a) + c2(a+b) = 0
Ko cần cảm ơn, mik giúp bạn chỉ vì mik đang sắp rơi vào danh sách học sinh dốt của hoc24h ^^
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
=\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)\)
mà a+b+c=0
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{0}{abc}\right)=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=9^2\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Rightarrow a^2+b^2+c^2\ge3\)
Lại có: \(a^2+b^2+c^2\ge ab+bc+ac\forall a,b,c\)
\(\Rightarrow3\ge ab+bc+ac\Rightarrow ab+bc+ac\le3\)
Bất đẳng thức ban đầu tương đương với:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{a\left(b^2+1\right)+b\left(c^2+1\right)+c\left(a^2+1\right)}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}a\left(b^2+1\right)\ge a\cdot2\sqrt{b^2}=2ba\\b\left(c^2+1\right)\ge b\cdot2\sqrt{c^2}=2cb\\c\left(a^2+1\right)\ge c\cdot2\sqrt{a^2}=2ac\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà \(ab+bc+ca\le3\)\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\cdot3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
\(VT=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\)
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) (1)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\)(2)
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c=1\)
Đặt biểu thức cần tìm GTNN là A .
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a+\dfrac{1}{4a}\text{≥}2\sqrt{a.\dfrac{1}{4a}}=1\)
\(b+\dfrac{1}{4b}\text{≥}2\sqrt{b.\dfrac{1}{4b}}=1\)
\(c+\dfrac{1}{4c}\text{≥}2\sqrt{c.\dfrac{1}{4c}}=1\) ≥
⇒ \(a+b+c+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}3\)
⇔ \(a+b+c+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}3+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ⇔ \(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{ ≥}3+\dfrac{3}{4}.\dfrac{\left(1+1+1\right)^2}{a+b+c}\text{ ≥}3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{15}{2}\)⇒ \(A_{Min}=\dfrac{15}{2}."="\text{⇔}a=b=c=\dfrac{1}{2}\)
bạn ơi làm thế nào để ra a+\(\dfrac{1}{4a}\)\(\ge\)2\(\sqrt{a\cdot\dfrac{1}{4a}}\) =1