Bất đẳng thức cần chứng minh tương đương với:

\(a^3b^2-a^2b^3+b^3c^2-c^3b^2+c^3a^2-c^2a^3\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b+b-a\right)\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+c^2a^2\left(b-a\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b\right)\ge0\)

\(\Leftrightarrow\left(a^2b^2-c^2a^2\right)\left(a-b\right)+\left(b^2c^2-c^2a^2\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow a^2\left(b^2-c^2\right)\left(a-b\right)+c^2\left(b^2-a^2\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left[a^2\left(b+c\right)-c^2\left(a+b\right)\right]\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left(a^2b+a^2c-c^2a-c^2b\right)\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left[a\left(ab-c^2\right)+c\left(a^2-bc\right)\right]\left(a-b\right)\left(b-c\right)\ge0\) luôn đúng do \(a\ge b\ge c\ge0\)

cảm ơn bạn nhá, bạn trả lời giúp mình mấy câu hỏi về BĐT còn lại của mik đc ko? cảm ơn bn nhiều!

Áp dụng bđt Cauchy-schwarz dạng engel ta có:

1. \(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{\left(a+b+c\right)^2}{\left(a+2b\right)+\left(b+2c\right)+\left(c+2a\right)}=\frac{a+b+c}{3}\)

Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)

2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)

Dấu "=" \(\Leftrightarrow a=b=c\)

\(P=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{2ac+3bc}\)

\(P\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{5\left(ab+bc+ca\right)}=\frac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

1 , a - ( a - b - c ) - ( b - c -a ) - ( c - b -a )

= a - a + b + c - b + c + a - c + b + a

= (a-a+a) + (b-b+b) + (c-c+c)

= a+b+c

2 , - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b - c - b + c + a + 1 - a - b - c + 3b

= (a+a-a) - (b+b+b) + (c-c+c) + 3b

= a - 3b + c + 3b

= a+c + (3b - 3b)

= a+c + 0

= a+c

3 , ( b - c - 6 ) - ( 7 - a + b ) + c

= b - c - 6 - 7 + a - b + c

= (b-b) + (c-c) - (6+7) + a

= 0 + 0 - 13 + a

= -13 + a

4 , - ( 3b - 2a - c ) - ( a - b - c ) - ( a - 2b -+ 2c )

= -3b + 2a + c - a + b + c - a  + 2b - 2c

= -3b + (2b + b) + (c + c) - (a+a) +2a - 2c

= -3b + 3b + 2c - 2a + 2a - 2c

= (3b - 3b) + (2c - 2c) + (2a + 2a)

= 0 + 0 + 0

= 0

chỉ bt lm đến đây thoy

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Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)

Theo hệ quả của BĐT AM-GM ta có:

\(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

1)    a - ( a - b - c ) - ( b - c - a ) - ( c - b - a )

= a - a + b + c - b + c + a - c + b + a

= 2a + b + c

2)  - ( a + b + c ) - ( b - c - a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b - c - b + c + a + 1 - a - b - c + 3b

= 1 - a - c

1,a-(a-b-c)-(b-c-a)-(c-b-a)

=a-a+b+c-b+c+a-c+b+a

=2a+b+c

2,-(a+b+c)-(b-c-a)+(1-a-b)-(c-3b)

=-a-b-c-b+c+a+1-a-b-c+3b

=1-a-c

3,(b-c-6)-(7-a+b)+c

=b-c-6-7+a-b+c

=a-13

4,-(3b-2a-c)-(a-b-c)-(a-2b+2c)

=-3b+2a+c-a+b+c-a+2b-2c

=0

5,(4a-3b+2c)-(4b-3c-2a)-(4c-3a+2b)+(a-b)-c

=4a-3b+2c-4b+3c+2a-4c+3a-2b+a-b-c

=(4a+2a+3a+a)-(3b+4b+2b+b)+(2c+3c-4c-c)

=10a-10b+0

=10.(a-b)

6,

2a-{a-b[a-b-(a+b+c)+2b]-c-b}

=2a-{a-b[a-b-a-b-c+2b]-c-b}

=2a-a-bc+c+b

=a-bc+c+b

=(a+b)-b(c-1)

a - ( a - b - c ) - ( b - c - a ) - ( c - b - a)

= a - a + b + c - b + c + a - c + b + a

= ( a -a + a ) + ( b - b + b ) + ( c + c - c)              ( vì mình ko có ngoặc vuông nên chỉ thế này thôi)

= a + b + c

Bạn tự làm hết nha

1)=>a-a+b+b-b+c+a-c+b+a=2a+2b+c=2(a+b)+c

2)=>-a-b-c-b+c+a+1-a-b-c+3b=-a

3)=>b-c-6-7+a-b+c=-13+a

4)-3b+2a+c-a+b+c-a+2b-2c=0

5)=>4a-3b+2c-4b+3c+2a-4c+3a-2b+a-b-c=-2a-10b-2c

 2a - { a - b [ a - b - ( a + b + c ) + 2b ] - c - b }

=2a-{a-b[a-b-a-b-c+2b]-c-b}

=2a-{a-bc-c-b}

=2a-a-bc-c-b

=a-bc-b-c