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Easy nà!
Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\) thì xyz = 1
BĐT trở thành: \(x^2+y^2+z^2\ge x+y+z\)
Áp dụng BĐT AM-GM,ta có: \(VT+1=\left(x^2+y^2\right)+\left(z^2+1\right)\)
\(\ge2xy+2z\ge2\sqrt{2xy.2z}=4\sqrt{xyz}=4\)
Suy ra \(VT\ge3\) (1)
Lại có: \(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
Cộng theo vế 3 BĐT: \(VT+3\ge2\left(x+y+z\right)\)
Kết hợp (1) suy ra \(2VT\ge VT+3\ge2\left(x+y+z\right)=2VP\)
Từ đây,ta có:\(2VT\ge2VP\Rightarrow VT\ge VP^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi x = y = z = 1
\(a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0.\)
\(a^2\left(\frac{a-}{b+c}\frac{b}{a+c}\right)+b^2\left(\frac{b}{a+c}\frac{-c}{a+b}\right)+c^2\left(\frac{c-}{a+b}\frac{a}{b+c}\right)\ge0.\)
\(a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)\ge0.\)
\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2.\) cái này dễ rồi .
\(\frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{bc^2+abc+ba^2}+\frac{c^2}{ca^2+abc+cb^2}\) (1)
Áp dụng BDT Cauchy-Schwarz: \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc}\)
Lại có: \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc=\left(ab+bc+ac\right)\left(a+b+c\right)\)
Thay vào -> dpcm
\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)
Áp dụng BĐT Cauchy dạng phân thức
\(\Rightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}\)
\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu "=" xảy ra khi a=b=c
Chúc bạn học tốt !!!
\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}\)
\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c\)
Chúc bạn học tốt !!!
Áp dụng bđt Cosi ta có: \(\frac{a^2}{a+b}+\frac{a+b}{4}\ge2;\frac{b^2}{b+c}+\frac{b+c}{4}\ge2;\frac{c^2}{c+d}+\frac{c+d}{4}\ge2\)\(;\frac{d^2}{d+a}+\frac{d+a}{4}\ge2\)
Cộng theo vế và a+b+c+d=1 ta có đpcm
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{a^2}{a+b}=\frac{a+b}{4};\frac{b^2}{b+c}=\frac{b+c}{4};\frac{c^2}{c+d}=\frac{c+d}{4};\frac{d^2}{d+a}=\frac{d+a}{4}\\\\a=b=c=1\end{cases}}\)
\(\Leftrightarrow a=b=c=d=\frac{1}{4}\)
sửa lại
\(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)
\(=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)
áp dụng bđt cauchy ta có:
\(b^2+1\ge2b;c^2+1\ge2c;a^2+1\ge2a\)
\(\Rightarrow a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\ge a-\frac{ab^2}{2b}+b-\frac{bc^2}{2b}+c-\frac{ca^2}{2a}\)
\(=a+b+c-\frac{ab+bc+ca}{2}\)
áp dụng cauchy ta có:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\left(Q.E.D\right)\)
dấu bằng xảy ra khi a=b=c=1
đặt \(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)
\(=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\le3-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)=3-\left(\frac{ab+bc+ca}{2}\right)\ge3-\frac{\left(a+b+c\right)^2}{6}=\frac{3}{2}\left(Q.E.D\right)\)
P/s: Không biết cách này có đúng không?
Chuyển vế qua và đặt thừa số chung,ta cần chứng minh:
\(a^2\left(\frac{1}{b+c}-\frac{1}{c+a}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0\)
\(\Leftrightarrow\frac{a^2\left(a-b\right)}{\left(b+c\right)\left(c+a\right)}+\frac{b^2\left(b-c\right)}{\left(a+c\right)\left(a+b\right)}+\frac{c^2\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}\ge0\)
\(\Leftrightarrow\frac{a^2\left(a-b\right)\left(a+b\right)+b^2\left(b-c\right)\left(b+c\right)+c^2\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\frac{a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)\ge0\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)
Đặt \(\left(a^2;b^2;c^2\right)\rightarrow\left(x;y;z\right)\).Ta cần chứng minh:
\(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)
Dấu "=" xảy ra khi x = y = z \(\Leftrightarrow a^2=b^2=c^2\Leftrightarrow a=b=c\)