=a-1

ko cần lời

\(P=\frac{a^2-1+1}{a-1}+\frac{2\left(b^2-1+1\right)}{b-1}+\frac{3\left(c^2-1+1\right)}{c-1}\)

\(P=a-1+2+\frac{1}{a-1}+2\left(b-1\right)+4+\frac{2}{b-1}+3\left(c-1\right)+6+\frac{3}{c-1}\)

=>\(P=a-1+\frac{1}{a-1}+2\left(b-1\right)+\frac{2}{b-1}+3\left(c-1\right)+\frac{3}{c-1}+12\)

ap dung bdt co si ta co

xay ra dau = khi va chi khi a=b=c=2

hình bạn đẹp

Đề bài là tìm MaxB 

Ta có \(a^2+b^2\ge2ab;b^2+1\ge2b\)

=> \(\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)

=> \(B\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)=\frac{1}{2}\)

Do \(abc=1\)=> \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}=1\)

MaxB=1/2  khi x=y=z=1

đặt a-1=x ; b-1=y; c-1=z (x,y,z>0)

\(P=\frac{\left(x+1\right)^2}{x}+\frac{2\left(y+1\right)^2}{y}+\frac{3\left(z+1\right)^2}{z}\)

\(=\frac{x^2+2x+1}{x}+\frac{2y^2+4y+2}{y}+\frac{3z^2+6z+3}{z}\)

\(=x+2+\frac{1}{x}+2y+4+\frac{2}{y}+3z+6+\frac{3}{z}\)

\(=\left(x+\frac{1}{x}\right)+\left(2y+\frac{2}{y}\right)+\left(3z+\frac{3}{z}\right)+12\)

Với x,y,z>0 áp dụng bđt AM-GM ta có: \(x+\frac{1}{x}\ge2\sqrt{x\cdot\frac{1}{x}}=2\)

\(2y+\frac{2}{y}\ge2\sqrt{2y\cdot\frac{2}{y}}=4;3z+\frac{3}{z}\ge2\sqrt{3z\cdot\frac{3}{z}}=6\)

Suy ra \(P\ge2+4+6+12=24\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{x}\\2y=\frac{2}{y}\\3z=\frac{3}{z}\end{matrix}\right.\Leftrightarrow x=y=z=1\Leftrightarrow a=b=c=2\)

\(P=\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(b+1\right)\left(c+1\right)}+\frac{c^3}{\left(c+1\right)\left(a+1\right)}-1\)

ôi trá hình :VVV

\(P=\frac{a^3}{\left(a+1\right).\left(b+1\right)}+\frac{b^3}{\left(b+1\right).\left(c+1\right)}+\frac{c^3}{\left(c+1\right).\left(a+1\right)}\)

Ko biết đúng hay không!

Mới lớp 6 , mà tôi nghĩ Lầy Văn Lội đúng đấy!

Ta có:

\(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+\left(b+c\right)+\left(b+c\right)}\)

\(\le\frac{1}{16}.\left(\frac{1}{a+b}+\frac{1}{c+a}+\frac{2}{b+c}\right)\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{1}{3a+2b+3c}\le\frac{1}{16}.\left(\frac{1}{b+c}+\frac{1}{a+b}+\frac{2}{c+a}\right)\left(2\right)\\\frac{1}{3a+3b+2c}\le\frac{1}{16}.\left(\frac{1}{c+a}+\frac{1}{b+c}+\frac{2}{a+b}\right)\left(3\right)\end{cases}}\)

Từ (1), (2), (3) \(\Rightarrow P\le\frac{1}{16}.\left(\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\right)\)

\(=\frac{1}{4}.2017=\frac{2017}{4}\)

đề thi vào lớp 10 năm nay của tỉnh thanh hóa

\(\frac{1}{a}\ge1-\frac{2}{2b+1}+1-\frac{3}{3c+2}=\frac{2b-1}{2b+1}+\frac{3c-1}{3c+2}\ge2\sqrt{\frac{\left(2b-1\right)\left(3c-1\right)}{\left(2b+1\right)\left(3c+2\right)}}\)

Tương tự: \(\frac{2}{2b+1}\ge\frac{a-1}{a}+\frac{3c-1}{3c+2}\ge2\sqrt{\frac{\left(a-1\right)\left(3c-1\right)}{a\left(3c+2\right)}}\)

\(\frac{3}{3c+2}\ge\frac{a-1}{a}+\frac{2b-1}{2b+1}\ge2\sqrt{\frac{\left(a-1\right)\left(2b-1\right)}{a\left(2b+1\right)}}\)

Nhân vế với vế:

\(\frac{6}{a\left(2b+1\right)\left(3c+2\right)}\ge\frac{8\left(a-1\right)\left(2b-1\right)\left(3c-1\right)}{a\left(2b+1\right)\left(3c+2\right)}\)

\(\Rightarrow\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\le\frac{3}{4}\)