Chứng minh các bất đẳng thức:
a/ aba+b+bcb+c+cac+aa+b+c2 đúng a>0,b>0,c>0
b/ aa+b+bb+c+cc+a<ab+c+bc+a+ca+b đúng a>0,b>0,c>0
c/ a>0,b>0:(1a)3+(ab)3+b31a+ab+b
d/ x>0,y>0,z>0:
2xx3+y2+2yy3+z2+2zz3+x21x2+1y2+1z2
a/  Xuất phát từ :(a+b)24ababa+ba+b4
Tương tự bcb+cb+c4;aca+ca+c4
Cộng từng vế aba+b+bcb+c+cac+aa+b+c2

b/  Với a>0;b>0;c>0:aa+b<a+ca+b+c
\frac{ a}{a+b} – \frac{ a+c}{a+b+c}= \frac{ a^{2} +ab +ac – a^{2} –ab –ac  -bc}{\left(  a+b  \right) \left(    a+b+c\right) }= \frac{ -bc}{\left(  a+b  \right) \left(    a+b+c\right) }<0
Tương tự \frac{ b}{b+c}< \frac{ a+b}{a+b+c}; \frac{ c}{a+c}< \frac{ b+c}{a+b+c}
Cộng từng vế: \frac{ a}{a+b}+ \frac{ b}{b+c}+ \frac{ c}{c+a} < \frac{ 2 \left(a+b+c    \right) }{a+b+c}=2 (1)
Bất đẳng thức Cosi áp dụng cho ab+c
a+ \left(   b+c \right)  \geq 2 \sqrt{ a \left( b+c   \right) } \Leftrightarrow \frac{ 1}{ \sqrt{ a \left( b+c   \right) }} \geq \frac{ 2}{\sqrt{a+b+c}}
\sqrt{ \frac{ s}{b+c}}= \frac{ a}{ \sqrt{ a \left( b+c   \right) }} \geq \frac{ 2}{a+b+c}
Tương tự \sqrt{ \frac{ b}{c+a}}lon \frac{ 2b}{a+b+c}; \sqrt{ \frac{ c}{a+b}} \geq \frac{ 2c}{a+b+c}
Cộng từng vế \sqrt{ \frac{ a}{b+c}}+ \sqrt{ \frac{ b}{c+a}}+ \sqrt{ \frac{ c}{a+b}} \geq \frac{ 2 \left( a+b+c   \right) }{a+b+c}=2
\frac{ a}{a+b}+ \frac{ b}{b+c}+ \frac{ c}{c+a} <2
\sqrt{ \frac{ a}{b+c}}+ \sqrt{ \frac{ b}{c+a}}+ \sqrt{ \frac{ c}{a+b}} \geq 2
\Rightarrow Đpcm.

c/ \left(\frac{ 1}{a}   \right)^{3} + \left( \frac{ a}{b}   \right)^{3} + b^{3} \geq \frac{ 1}{a}+ \frac{ a}{b}+b
Áp dụng bất đẳng thức Cosi cho ba số dương:
\left(  \frac{ 1}{a}  \right)^{3}+1+1 \geq 3 \sqrt[ 3]{ \frac{ 1}{a^{3}}}=3.\frac{ 1}{3}; \left( \frac{ a}{b}   \right)^{3} +1+1 \geq 3 \sqrt[3]{ \left(  \frac{ a}{b}  \right)^{2} }=3. \frac{ a}{b}
b^{3}+1+1 \geq 3 \sqrt[3]{b^{3}} =3b
\Rightarrow \left(\frac{ 1}{a}   \right)^{3} + \left( \frac{ a}{b}   \right)^{3} + b^{3} +6 \geq 3 \left( \frac{ 1}{a}+ \frac{ a}{b}+ b   \right) = \frac{ 1}{a}+ \frac{ a}{b}+b+2 \left(\frac{ 1}{a}+ \frac{ a}{b}+b    \right)
Nhưng 2 \left(   \frac{ 1}{a}+ \frac{ a}{b}+b \right)  \geq 2. 3 \sqrt[3]{ \frac{ 1}{a}. \frac{ a}{b}.b} =6
\Rightarrow \left(\frac{ 1}{a}   \right)^{3} + \left( \frac{ a}{b}   \right)^{3} + b^{3}+b \geq \frac{ 1}{a}+ \frac{ a}{b}+b + 2 \left(\frac{ 1}{a}+ \frac{ a}{b}+b   \right) \geq \frac{ 1}{a}+ \frac{ a}{b}+b +6
\Leftrightarrow \left( - \frac{ 1}{a}   \right)^{2} + \left(\frac{ a}{b}    \right)^{2} +b^{3} \geq \frac{ 1}{a}+ \frac{ a}{b}+b

Có đẳng thức khi a=b=1
d/  Áp dụng bất đẳng thức Cosi:
x^{2} +y^{2} \geq 2 \sqrt{ x^{2} y^{2}}= 2xy \sqrt{ x} \Leftrightarrow \frac{ 2 \sqrt{ x}}{x^{3}+y^{2}} \leq \frac{ 1}{xy}
Tương tự: \frac{ 2 \sqrt{ y}}{y^{3}+z^{2}} \leq \frac{ 1}{yz}; \frac{ 2 \sqrt{ z}}{y^{3}+x^{2}} \leq \frac{ 1}{zx}
Cộng từng vế: \frac{ 2 \sqrt{ x}}{x^{3}+y^{2}} \leq \frac{ 1}{xy}+\frac{ 2 \sqrt{ y}}{y^{3}+z^{2}} +\frac{ 2 \sqrt{ z}}{y^{3}+x^{2}} \leq \frac{ 1}{xy}+\frac{ 1}{yz}+\frac{ 1}{zx}
Nhưng  \begin{cases}  \frac{ 1}{xy} \leq \frac{ 1}{2} \left(  \frac{ 1}{ x^{2} }+ \frac{ 1}{y^{2}}   \right) \\ \frac{ 1}{yz} \leq \frac{ 1}{2} \left( \frac{ 1}{y^{2}} + \frac{ 1}{z^{2}}   \right)   \\   \frac{ y}{ z^{2}} \leq \frac{ 1}{2} \left(  \frac{ 1}{z^{2}} \frac{ 1}{ x^{2} }  \right)   \end{cases} \Rightarrow \frac{ 1}{xy}+\frac{ 1}{yz}+\frac{ 1}{zx} \leq \frac{ 1}{  x^{2} }+ \frac{ 1}{y^{2}}+ \frac{ 1}{z^{2}} đpcm

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