a/ Bạn cứ khai triển biến đổi tương đương thôi (mà làm biếng lắm)

b/ Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)

\(VT=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

\(VT\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)\ge\frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Áp dụng Buhiacopxki có \(\left(\left(\frac{m}{\sqrt{x}}\right)^2+\left(\frac{n}{\sqrt{y}}\right)^2\right)\left(\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right)\ge\left(m+n\right)^2\)

\(\RightarrowĐPCM\)

b) với mọi a,b,c ϵ R và x,y,z ≥ 0 có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(1\right)\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Thật vậy với a,b∈ R và x,y ≥ 0 ta có:
\(\frac{a^2}{x}=\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(2\right)\)
⇔\(\frac{a^2y}{xy}+\frac{b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}.\left(x+y\right)xy\ge\frac{\left(a+b\right)^2}{x+y}.\left(x+y\right)xy\)
⇔\(\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2xy\)
⇔\(a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+2abxy+b^2xy\)
⇔\(b^2x^2+a^2y^2-2abxy\ge0\)
⇔\(\left(bx-ay\right)^2\ge0\)(luôn đúng )
Áp dụng BĐT (2) có:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)} \)
= \(\frac{1}{a^2}.\frac{1}{ab+ac}+\frac{1}{b^2}.\frac{1}{bc+ac}+\frac{1}{c^2}.\frac{1}{ac+bc}\)
=\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)
Áp dụng BĐT (1) ta có:
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}++\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}\)
Mà abc=1⇒\(\left\{{}\begin{matrix}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{matrix}\right.\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\sqrt[3]{\frac{1}{1}}=3\)( BĐT cosi )
⇒\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
⇒\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}.3=\frac{3}{2}\)
Vậy \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Chúc bạn học tốt !!!

Ta có: abc = 1, thế vào ta được:

\(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

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

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

Áp dụng BĐT Cauchy - Schwarz dạng Engel, ta có:

\(VT\ge\frac{\left(bc+ca+ac\right)^2}{abc\left(2ab+2bc+2ca\right)}=\frac{\left(bc+ca+ac\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)

Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)

Đặt vế trái là P

Ta có: \(P=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Bài 1 : \(VT=a^2+b^2+c^2+3abc=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)+3abc\left(a+b+c\right)}{a+b+c}\ge\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc}{a+b+c}\)

\(=\frac{a^3+b^3+c^3+3abc+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+6abc}{a+b+c}\)

\(\ge\frac{2ab\left(a+b\right)+2bc\left(b+c\right)+2ca\left(c+a\right)+6abc}{a+b+c}\)

\(=\frac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{a+b+c}=6\)

Có sai sót gì xin cmt bên dưới ạ

Mình xin đưa ra lời giải

2/ Áp dụng BĐT Cauchy Schwarz dạng Engel:\(\text{VT}= \sum \frac{a^2 (\frac{7}{2}a +c)^2}{(\frac{7}{2}a +c)^2 \Big[(a+b+c)a+b^2\Big]} \geqq \frac{\Big[\sum a ( \frac{7}{2}a +c)\Big]^2}{\sum (\frac{7}{2}a +c)^2 \Big[(a+b+c)a+b^2\Big]}\geqq \frac{3}{4}\)

Bất đẳng thức cuối đúng, tương đối đơn giản.

3/ Giả sử $a=\max\{a,b,c\}$. Ta có:\(\text{LHS -RHS} = {\frac { \left( a-b \right) ^{2} \left( 8\,c+15\,a+31\,b \right) }{ 2592}}+{\frac { \left( -2\,c+a+b \right) ^{2} \left( c-a+b \right) }{ 144}}+{\frac { \left( -6\,c-a+7\,b \right) ^{2} \left( -2\,c+3\,a-b \right) }{2592}} \geqq 0 \)4/\(LHS^2-RHS^2=\frac{1}{4}\sum\frac{a\left(12ac+3b^2+4bc\right)\left(b-c\right)^2}{bc^2\left(ab+bc+ca\right)}+\frac{1}{4}\sum\frac{a\left(2ac-b^2-bc\right)^2}{bc^2\left(ab+bc+ca\right)}\ge0\)

5/ Dùng Buffalo Way, giả sử \(c=\min\{a,b,c\}\) và đặt \(a=c+u,b=c+v\)

Đơn giản thế mà ít người tham gia.

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=2\)

\(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2b}{b}}=2a\) ; \(\frac{b^2}{c}+c\ge2b\) ; \(\frac{c^2}{a}+a\ge2a\)

Cộng vế với vế:

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2\left(a+b+c\right)\)

\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c=6\)

Dấu "=" xảy ra khi \(a=b=c=2\)

\(P\left(x\right)=48x^3-24x^2+3x+16x^2-8x+1\)

\(=3x\left(16x^2-8x+1\right)+16x^2-8x+1\)

\(=\left(3x+1\right)\left(16x^2-8x+1\right)\)

\(=\left(3x+1\right)\left(4x-1\right)^2\)

b/ \(\Leftrightarrow48x^3-8x^2\ge5x-1\)

\(\Leftrightarrow48x^3-8x^2-5x+1\ge0\)

\(\Leftrightarrow\left(3x+1\right)\left(4x-1\right)^2\ge0\) (luôn đúng \(\forall x\ge0\))

Dấu "=" xảy ra khi \(x=\frac{1}{4}\)

c/ Bạn chắc là ghi đề sai

Đặt vế trái là P

\(a^3+a^3+\frac{1}{64}\ge3\sqrt[3]{\frac{a^6}{64}}=\frac{3}{4}a^2\)

\(\Rightarrow6a^3+\frac{3}{64}\ge\frac{9}{4}a^2\)

Tương tự ta có: \(6b^3+\frac{3}{64}\ge\frac{9}{4}b^2\) ; \(6c^3+\frac{3}{64}\ge\frac{9}{4}c^2\) ; \(6d^2+\frac{3}{64}\ge\frac{9}{4}d^2\)

Cộng vế với vế:

\(6\left(a^3+b^3+c^3+d^3\right)+\frac{12}{64}\ge\frac{9}{4}\left(a^2+b^2+c^2+d^2\right)\)

\(\Rightarrow P=6\left(a^3+b^3+c^3+d^3\right)-\left(a^2+b^2+c^2+d^2\right)\ge\frac{5}{4}\left(a^2+b^2+c^2+d^2\right)-\frac{12}{64}\)

Mà \(a^2+b^2+c^2+d^2\ge\frac{\left(a+b+c+d\right)^2}{4}\)

\(\Rightarrow P\ge\frac{5}{16}\left(a+b+c+d\right)^2-\frac{12}{64}=\frac{1}{8}\)

Dấu "=" xảy ra khi \(a=b=c=d=\frac{1}{4}\)