Từ \(2a+2b+2c=3abc\)

\(\Leftrightarrow\frac{2}{3bc}+\frac{2}{3ac}+\frac{2}{3ab}=1\left(1\right)\)

Khi đó \(P=\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}\)

Áp dụng BĐT AM-GM ta có: 

\(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}\ge3\sqrt[3]{\frac{b}{a^2}\cdot\frac{c}{b^2}\cdot\frac{a}{c^2}}=3\sqrt[3]{\frac{1}{abc}}\)

\(P_{Min}\) xảy ra khi \(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}=3\sqrt[3]{\frac{1}{abc}}\forall a=b=c\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow a=b=c=\sqrt{2}\)

Khi đó \(P_{Min}=3\sqrt[3]{\frac{1}{abc}}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}=\frac{3\sqrt{2}-6}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\sqrt{2}\)

Bài này giải như này cơ:

\(2a+2b+2c=3abc\)\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{2}\)

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

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

\(\ge\frac{2\left(a-1\right)}{ac}+\frac{2\left(b-1\right)}{ab}+\frac{2\left(c-1\right)}{bc}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

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

\(\ge\sqrt{3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}-3=\sqrt{3.\frac{3}{2}}-3=\frac{3\sqrt{2}-6}{2}\)

Vậy \(minP=\frac{3\sqrt{2}-6}{2}\Leftrightarrow a=b=c=\sqrt{2}\)

cả 2 cách đều hay :*

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

qua học 24 mà coi

\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)

xong AM -GM

=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

sao toàn toán lớp 9 thế

\(a-\frac{ab^2}{b^2+1}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự và cộng lại, ta có:\(p\ge a+b+c-\frac{ab+bc+ca}{2}\) mà 3(ab+bc+ca)\(\le\)(a+b+c)^2=9

=>ab+bc+ca\(\le\)3

=> \(p\ge3-\frac{3}{2}=\frac{3}{2}\)

Dấu = xảy ra =>a=b=c=1

Vậy còn cách tìm maxP thì sao hả mấy bạn