áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
\(\frac{a^3}{a^2+b^2}=\frac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\frac{ab^2}{a^2+b^2}\ge a-\frac{ab^2}{2ab}=a-\frac{b}{2}\)(cosi ngược dấu)
Chứng minh tương tự được
\(\frac{b^3}{b^2+c^2}\ge b-\frac{c}{2};\frac{c^3}{c^2+a^2}\ge c-\frac{a}{2}\)
Cộng các vế của 3 bất đẳng thức
\(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\ge a+b+c-\frac{a+b+c}{2}=\frac{a+b+c}{2}\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a-\frac{a^2}{a+b}+b-\frac{b^2}{b+c}+c-\frac{c^2}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
\(\Leftrightarrow a^2\left(a+b\right)\left(a+c\right)+b^2\left(b+a\right)\left(b+c\right)+c^2\left(c+a\right)\left(c+b\right)\ge a^2\left(a+c\right)\left(b+c\right)+b^2\left(b+a\right)\left(c+a\right)+c^2\left(c+b\right)\left(a+b\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2c^2+a^2b^2+b^2c^2\left(lđ\right)\)
\(\Leftrightarrow\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\ge a+b+c\)
CMR:
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\)
\(\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\)
\(\)
cho a,b,c thỏa mãn: \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{c^2}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}\ge\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)
thì \(|a|=|b|=|c|\)
Chứng minh \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
C1\(VT=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2=VP\)
Dấu bằng xảy khi a=b=c
C2 Áp dụng cosi ta có :
\(\frac{a^3}{b}+\frac{a^3}{b}+b^2\ge3a^2\);\(\frac{b^3}{c}+\frac{b^3}{c}+c^2\ge3b^2\); \(\frac{c^3}{a}+\frac{c^3}{a}+a^2\ge3c^2\)
Cộng 3 vế của 3 BĐT ta được ĐPCM
Cho a, b, c > 0. CMR:
a, \(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge2\left(a+b+c\right)\)
b, \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
c, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
Giúp mình với các bạn ơiii
1) Cho a, b, c > 0. Chứng minh: \(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
2) Cho \(a,b,c\in R\).
a) Chứng minh: \(\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a+b+c+1\right)^2\)
b) Chứng minh: \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
3) Cho \(a,b,c\in R\)Chứng minh: \(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
Chứng minh rằng nếu a,b,c là các số thỏa mãn các bấc đẳng thức sau:\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{c^2}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}\ge\frac{b^2}{b+a}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)
Thì \(|a|=|b|=|c|\)
CMR:với mọi số thực dương a;b;c ta có:
\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)