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2. \(BĐT\Leftrightarrow\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\ge1\)
Đặt\(\frac{2}{a}=x;\frac{2}{b}=y;\frac{2}{c}=z\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=8\end{cases}}\)
Ta cần chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge1\Leftrightarrow\left(yz+y+z+1\right)+\left(zx+z+x+1\right)+\left(xy+x+y+1\right)\ge xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\)\(\Leftrightarrow x+y+z\ge6\)(Đúng vì \(x+y+z\ge3\sqrt[3]{xyz}=6\))
Đẳng thức xảy ra khi x = y = z = 2 hay a = b = c = 1
3. Ta có: \(a+b+c\le\sqrt{3}\Rightarrow\left(a+b+c\right)^2\le3\)
Ta có đánh giá quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Từ đó suy ra \(ab+bc+ca\le1\)
\(A=\frac{\sqrt{a^2+1}}{b+c}+\frac{\sqrt{b^2+1}}{c+a}+\frac{\sqrt{c^2+1}}{a+b}\ge\frac{\sqrt{a^2+ab+bc+ca}}{b+c}+\frac{\sqrt{b^2+ab+bc+ca}}{c+a}+\frac{\sqrt{c^2+ab+bc+ca}}{a+b}\)\(=\frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}+\frac{\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}+\frac{\sqrt{\left(c+a\right)\left(c+b\right)}}{a+b}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=3\)Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
\(\frac{a^2}{1+b}+\frac{1+b}{4}\ge a\)
\(\frac{b^2}{1+c}+\frac{1+c}{4}\ge b\)
\(\frac{c^2}{1+a}+\frac{1+a}{4}\ge c\)
=>\(A\ge a+b+c-\frac{1}{4}\left(3+a+b+c\right)=\frac{3}{4}\left(a+b+c-1\right)\ge\frac{3}{4}\left(3\sqrt[3]{abc}-1\right)=\frac{3}{2}\)
A min = 3/2 khi x= y =z =1
\(\frac{a^2}{1+b}=\frac{a^2\left(1+b\right)-a^2b}{1+b}=a^2-\frac{a^2b}{1+b}\ge a^2-\frac{a^2b}{2\sqrt{b}}=a^2-\frac{a^2\sqrt{b}}{2}\) và tương tự
Ta có:
\(\frac{a+1}{1+b^2}=a+1-\frac{\left(a+1\right)b^2}{1+b^2}\ge a+1-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\left(1\right)\)
Tương tụ ta có:
\(\hept{\begin{cases}\frac{\left(b+1\right)}{1+c^2}\ge b+1-\frac{bc+c}{2}\left(2\right)\\\frac{\left(c+1\right)}{1+a^2}\ge c+1-\frac{ca+a}{2}\left(3\right)\end{cases}}\)
Từ (1), (2), (3) ta có:
\(M\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(=3+3-\frac{ab+bc+ca+3}{2}\)
\(\ge\frac{9}{2}-\frac{\left(a+b+c\right)^2}{6}=3\)
Dùng bđt AM - GM cho 7 số; 2 số và 3 số không âm, ta được:
\(a^3c^2+a^3c^2+a^3c^2+b^3a^2+b^3a^2+1+1\ge7a\)(1)
\(b^3a^2+b^3a^2+b^3a^2+c^3b^2+c^3b^2+1+1\ge7b\)(2)
\(c^3b^2+c^3b^2+c^3b^2+a^3c^2+a^3c^2+1+1\ge7c\)(3)
\(\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}\ge3\)
\(a+b+c\ge3\)
Từ (1); (2); (3) suy ra \(a^3c^2+b^3a^2+c^3b^2\ge\frac{7\left(a+b+c\right)}{5}-\frac{6}{5}\)
\(P=\text{Σ}_{cyc}\frac{a}{b^2}+\frac{9}{2\left(a+b+c\right)}=\text{Σ}_{cyc}a^3c^2+\frac{9}{2\left(a+b+c\right)}\)
\(\ge\frac{7\left(a+b+c\right)}{5}+\frac{9}{2\left(a+b+c\right)}-\frac{6}{5}\)
\(=\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}+\frac{9\left(a+b+c\right)}{10}-\frac{6}{5}\)
\(\ge3+\frac{9}{10}.3-\frac{6}{5}=\frac{9}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
\(A=\frac{1}{a^2\left(b+c\right)}+\frac{1}{b^2\left(c+a\right)}+\frac{1}{c^2\left(a+b\right)}\)
\(=\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\)
\(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)
Đặt: \(ab=x;bc=y;ac=z\)=> xyz = 1; x,y,z>0
\(A=\frac{y}{x+z}+\frac{z}{y+x}+\frac{x}{z+y}=\frac{y^2}{xy+yz}+\frac{z^2}{yz+xz}+\frac{x^2}{zx+xy}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+xz+xz\right)}\ge\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\frac{3}{2}\)
Dấu "=" xảy ra <=> x = y = z= 1 => a = b = c = 1
Vậy gtnn của A = 3/2 tại a = b = c = 1
