Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\dfrac{b}{a}=x;\dfrac{c}{b}=y\).
Ta có: \(P=\dfrac{1}{\left(\dfrac{a+b}{a}\right)^2}+\dfrac{1}{\left(\dfrac{b+c}{b}\right)^2}+\dfrac{b}{a}.\dfrac{c}{b}.\dfrac{1}{4}\)
\(P=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}+\dfrac{xy}{4}\).
Ta có bđt quen thuộc: \(\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}\ge\dfrac{1}{xy+1}\) (bạn xem cm ở đây).
Do đó \(P\ge\dfrac{1}{xy+1}+\dfrac{xy+1}{4}-\dfrac{1}{4}\ge1-\dfrac{1}{4}=\dfrac{3}{4}\).
Đẳng thức xảy ra khi x = y = 1 tức a = b = c.
Vậy...
BĐT phụ kia có 1 cách chứng minh rất hay mà không cần đến biến đổi tương đương với mũ to:
\(\dfrac{1}{\left(1.1+\sqrt{xy}.\sqrt{\dfrac{x}{y}}\right)^2}+\dfrac{1}{\left(1.1+\sqrt{xy}.\sqrt{\dfrac{y}{x}}\right)^2}\ge\dfrac{1}{\left(1+xy\right)\left(1+\dfrac{x}{y}\right)}+\dfrac{1}{\left(1+xy\right)\left(1+\dfrac{y}{x}\right)}=\dfrac{1}{1+xy}\)
Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).
Áp dụng bđt Schwars ta có:
\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).
Đẳng thức xảy ra khi a = b = c = 1.
\(P=\dfrac{a^4}{a^2b^2+a^2c^4}+\dfrac{b^4}{b^2c^2+a^2b^2}+\dfrac{c^4}{a^2+b^2}-\dfrac{12abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(P\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}-\dfrac{12abc}{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}}\)
\(P\ge\dfrac{3\left(a^2b^2+b^2c^2+c^2a^2\right)}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}-\dfrac{3}{2}=0\)
\(P_{min}=0\) khi \(a=b=c\)
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^
\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\right)\)
\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\right)\)
\(P\ge\left(a+b+c\right)^2\left(\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{\dfrac{1}{3}\left(a+b+c\right)^2}\right)=30\)
\(P_{min}=30\) khi \(a=b=c\)
\(M=\dfrac{\left(ab\right)^2}{abc^2\left(a+b\right)}+\dfrac{\left(ac\right)^2}{acb^2\left(a+c\right)}+\dfrac{\left(bc\right)^2}{a^2bc\left(b+c\right)}\)
\(M\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2abc}=\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{6abc}\ge\dfrac{9abc}{6abc}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)}{a^2c^2+2ab^2c}\)
\(P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)
\(P\ge\dfrac{\left[a^2+b^2+c^2+3abc\right]^2}{\left(ab+bc+ca\right)^2}\)
Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)
Ta có: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)
\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)
\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)
sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2
Tử, mẫu không đồng bậc
Đề sai hoặc thiếu điều kiện
tử cộng thêm c^2 bớts c^2
tách tử theo mẫu
cô si mẫu