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a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)
\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)
\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)
Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)
\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)
Rồi tương tự các kiểu:v
Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)
\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))
Không phải dùng tới Cauchy-Schwarz:D
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(A\ge\dfrac{\left(1+1+1\right)^2}{3+ab+bc+ac}=\dfrac{9}{3+ab+bc+ac}\)
Mặt khác,theo hệ quả AM-GM: \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}\le\dfrac{3^2}{3}=3\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ac}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Bạn học delta chưa nhỉ, HSG chắc chắn là học rồi:vv
delta là độ chênh lệch đó hả???肖赵战颖