Chủ đề này có 3 trả lời

[yeutoanmaimai1]

1, Cho a,b,c đôi một khác nhau. C/m

$\frac{a^{3}-b^{3}}{(a-b)^{3}}+\frac{b^{3}-c^{3}}{(b-c)^{3}}+\frac{c^{3}-a^{3}}{(c-a)^{3}}$

2, cho $a,b,c\geq 0$ Cm

a, $\sqrt{a^{2}+ab+b^{2}}+\sqrt{b^{2}+bc+c^{2}}+\sqrt{c^{2}+ca+a^{2}}\geq \sqrt{3}(a+b+c)$

b, $\sum a(-\sqrt{a}+\sqrt{b}+\sqrt{c})\leq 3\sqrt{abc}$

3, cho a,b,c>0 Cm

$\frac{a^{8}+b^{8}+c^{8}}{a^{3}*b^{3}*c^{3}}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

4, cho a,b,c,d>0 Cm

$\frac{a+b}{a+b+c}+\frac{b+c}{b+c+d}+\frac{c+d}{c+d+a}+\frac{d+a}{d+a+b}< 3$

[rainbow99]

4, cho a,b,c,d>0 Cm

$\frac{a+b}{a+b+c}+\frac{b+c}{b+c+d}+\frac{c+d}{c+d+a}+\frac{d+a}{d+a+b}< 3$

Vì a,b,c,d>0 nên ta có:$\frac{a+b}{a+b+c}<\frac{a+b+d}{a+b+c+d}$

Tương tự: $\frac{b+c}{b+c+d}<\frac{b+c+a}{b+c+d+a}$

                $\frac{c+d}{c+d+a}<\frac{c+d+b}{a+b+c+d}$

                 $\frac{d+a}{d+a+b}<\frac{d+a+c}{a+b+c+d}$

Cộng các bất đẳng thức trên lại với nhau ta có bđt cần chứng minh

[rainbow99]

2, cho $a,b,c\geq 0$ Cm

a, $\sqrt{a^{2}+ab+b^{2}}+\sqrt{b^{2}+bc+c^{2}}+\sqrt{c^{2}+ca+a^{2}}\geq \sqrt{3}(a+b+c)$

Ta có $\sqrt{x^{2}+xy+y^{2}}+\sqrt{y^{2}+yz+z^{2}}+\sqrt{x^{2}+xz+z^{2}}=\sqrt{(x+\frac{1}{2}y)^{2}+\frac{3}{4}y^{2}}+\sqrt{(y+\frac{1}{2}z)^{2}+\frac{3}{4}z^{2}}+\sqrt{(z+\frac{1}{2}x)^{2}+\frac{3}{4}x^{2}}$$\geq \sqrt{(x+\frac{1}{2}y+y+\frac{1}{2}z+z+\frac{1}{2}x)^{2}+\frac{3}{4}(x+y+z)^{2}}=\sqrt{3}(x+y+z)$

[huykinhcan99]

3, cho a,b,c>0 Cm

$\frac{a^{8}+b^{8}+c^{8}}{a^{3}*b^{3}*c^{3}}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

 

Sử dụng bất đẳng thức $AM - GM$ ta có:

$$a^8+a^8+b^8+b^8+c^8+c^8+c^8\geqslant 8\sqrt[8]{a^{16}b^{24}c^{24}}$$

\begin{equation} \label{eq:1} \Leftrightarrow 2a^8+3b^8+3c^8\geqslant 8a^2b^3c^3 \end{equation}

 

Tương tự:

\begin{align} \label{eq:2} 3a^8+2b^8+3c^8\geqslant 8a^3b^2c^3 \\ \label{eq:3} 3a^8+3b^8+2c^8\geqslant 8a^3b^3c^2 \end{align}

 

Cộng \eqref{eq:1}, \eqref{eq:2}, \eqref{eq:3} vế với vế ta thu được:

\begin{align*} &a^8+b^8+c^8 \geqslant a^2b^3c^3+a^3b^2c^3+a^3b^3c^2 \\ \Leftrightarrow \ & a^8+b^8+c^8 \geqslant a^3b^3c^3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) \\ \Leftrightarrow \ & \dfrac{a^8+b^8+c^8}{a^3b^3c^3} \geqslant \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \tag{$\blacksquare$}\end{align*}