Áp dụng Bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có:

\(P\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)

Lại có:

\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\)

\(\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=9\)

Mặt khác \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)

\(\Rightarrow\frac{1}{ab+bc+ca}\ge3\)\(\Rightarrow P_{Min}=30\)

Dấu = khi \(a=b=c=\frac{1}{3}\)

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

\(\ge4+4-3=5\)

\(\frac{a}{b}+\frac{b}{a}=\frac{a^2}{ab}+\frac{b^2}{ab}=\frac{a^2+b^2}{ab}\ge2\)

Vậy Min A = 2 \(\Leftrightarrow a=b\)

a2(b+c)2+5bc+b2(a+c)2+5ac≥4a29(b+c)2+4b29(a+c)2=49(a2(1−a)2+b2(1−b)2)(vì a+b+c=1)
a2(1−a)2−9a−24=(2−x)(3x−1)24(1−a)2≥0(vì )<a<1)
⇒a2(1−a)2≥9a−24
tương tự: b2(1−b)2≥9b−24
⇒P⩾49(9a−24+9b−24)−3(a+b)24=(a+b)−94−3(a+b)24.
đặt t=a+b(0<t<1)⇒P≥F(t)=−3t24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13

B=1^8trên1^2

\(\frac{1}{12}\)

Ta có: 

\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2}{\frac{1}{a}+\frac{1}{c}}=\frac{2ac}{a+c}\)

Thế \(b=\frac{2ac}{a+c}\) vào M, ta được:

 \(M=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{1+\frac{2c}{a+c}}{2-\frac{2c}{a+c}}+\frac{1+\frac{2a}{a+c}}{2-\frac{2a}{a+c}}\)

\(M=\frac{\left(a+c\right)+2c}{2\left(a+c\right)-2c}+\frac{\left(a+c\right)+2a}{2\left(a+c\right)-2a}=\frac{a+3c}{2a}+\frac{3a+c}{2c}\)

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

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

Xét \(\frac{a}{c}+\frac{c}{a}\ge2\Leftrightarrow...\)(bạn tự biến đổi tương đương để chứng minh nó nhé)

(ĐK xảy ra dấu "=": a=c)

Do đó \(M+2=\frac{3}{2}\left(1+\frac{a}{c}+\frac{c}{a}+1\right)=\frac{3}{2}\left(2+\frac{a}{c}+\frac{c}{a}\right)\ge\frac{3}{2}\left(2+2\right)=6\Leftrightarrow M\ge4\)

Vậy GTNN của \(M=4\)khi \(a=c\Leftrightarrow\frac{2}{b}=\frac{2}{a}\Leftrightarrow b=a=c\)

Chúc bạn học tốt!

P/S: bài này khó thật đấy! Mình chuyên toán 9 mà giải hết nửa tiếng mới xong :D!