Question

[SA Na]

giúp mình vài câu sau nha
thanks nhiều

1. cho a,b,c > 0
C/m: \(\dfrac{a^3+b^3}{ab}+\dfrac{b^3+c^3}{bc}+\dfrac{c^3+a^3}{ca}>=2\left(a+b+c\right)\)

2. cho a,b,c > 0
C/m: \(\dfrac{a^4}{bc^2}+\dfrac{b^4}{ca^2}+\dfrac{c^4}{ab^2}>=a+b+c\)

3. cho a,b,c > 0 và \(a^2+b^2+c^2=3\)
C/m: \(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}>=\dfrac{3}{2}\)

4. cho a,b,c > 0
C/m: \(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}>=\dfrac{a+b+c}{2}\)

Answer 1

[Nguyễn Huy Tú]

Bài 3:

Ta có: \(a^2+b^2+c^2=3\ge ab+bc+ca\) ( tự cm bđt nha )

Áp dụng bất đẳng thức Schwarz ta có:

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+bc}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{9}{6}=\dfrac{3}{2}\)

\(\Rightarrowđpcm\)

Dấu " = " khi a = b = c = 1

Bài 4:

Ta có: \(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)

( BĐT AM - GM )

Tương tự \(\Rightarrow\dfrac{b^3}{c^2+a^2}\ge b-\dfrac{c}{2}\)

\(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)

\(\Rightarrow VT\ge\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{2}\)

Dấu " = " khi a = b = c

Answer 2

[Lightning Farron]

Tiếp sức cho Tú đệ

Bài 1: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

\(\Rightarrow\dfrac{a^3+b^3}{ab}\ge\dfrac{ab\left(a+b\right)}{ab}=a+b\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\ge VP."="\Leftrightarrow a=b=c\)

Bài 2: Holder:

\(\left(\dfrac{a^4}{bc^2}+\dfrac{b^4}{ca^2}+\dfrac{c^4}{ab^2}\right)\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)\left(c+a+b\right)\ge\left(a+b+c\right)^3\)

Cần chứng minh \(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\ge a+b+c\)

AM-GM: \(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{bc}{a}\cdot\dfrac{ca}{b}}=2c\)

Tương tự rồi cộng theo vế:

\("=" \Leftrightarrow a=b=c\)