Sangnguyen3

$\Leftrightarrow \sum \frac{2a}{b+c} -3\leq 2\left ( \sum \frac{a}{b} -3\right )$

$2\left ( \sum \frac{a}{b+c} -\frac{3}{2} \right )\leq 2\left ( \sum \frac{a}{b} -3\right )$

$\sum \frac{a}{b}-3=\sum \frac{a^{2}}{ab}-3\geq \frac{\left ( \sum a \right )^{2}}{\sum ab}-3=\frac{\sum (a-b)^{2}}{2\sum ab}$

$\Rightarrow 2\left ( \sum \frac{a}{b}-3 \right )\geq \frac{\sum (a-b)^{2}}{\sum ab}$

$2\left ( \sum \frac{a}{b+c}-\frac{3}{2} \right )=2\left ( \sum \frac{(a-b)^{2}}{2(a+c)(b+c)} \right )=\sum \frac{(a-b)^{2}}{(a+c)(b+c)}$

Can : $\sum \frac{(a-b)^{2}}{(a+c)(b+c)}\leq \sum \frac{(a-b)^{2}}{ab+bc+ca}$

$\Leftrightarrow \sum \left ( a-b\right )^{2}\left ( \frac{1}{\sum ab} -\frac{1}{(a+c)(b+c)}\right )\geq 0 \Leftrightarrow\sum \left ( a-b \right )^{2}\left ( \frac{c^{2}}{\sum ab.(a+c)(b+c)} \right )\geq 0$

$\left\{\begin{matrix} p-1=2x(x+2) (1) & & \\ p^{2}-1=2y(y+2) (2)& & \end{matrix}\right.$

Lay (2)-(1) , ta co :

$p^{2}-p=2\left ( y-x \right )\left ( y+x \right ) \Leftrightarrow p(p-1)=2\left ( y-x \right )\left ( y+x \right )$

$x< p,y< p\Rightarrow 0< y-x< p \Rightarrow \left ( y-x;p \right )=1$

$p=2 \Rightarrow 2x(x+2)=1 (VL)$

$\Rightarrow p> 2 \Rightarrow \left ( 2;p \right )=1$

$2\left ( y-x \right )\left ( y+x \right )\vdots p\Rightarrow y+x\vdots p$

$0

The vao giai tim dc x,y

Bổ đề :  $a+b+c+abc=4$ thì $a+b+c\geq ab+bc+ca$

Chứng minh : https://diendantoanh...g-abcgeqabbcca/

$\sum \frac{a}{\sqrt{b+c}}=\sum \frac{a^{2}}{\sqrt{a}.\sqrt{ab+ac}}\geq \frac{\left ( \sum a \right )^{2}}{\sum \left (\sqrt{a}.\sqrt{ab+ac} \right ) }\geq \frac{\left ( \sum a \right )^{2}}{\sqrt{2(a+b+c)(ab+bc+ca)}}\geq \frac{\left ( \sum a \right )^{2}}{\sqrt{2}(a+b+c)}=\frac{\sqrt{2}}{2}(a+b+c)$

$a^{2}-a+1=\left ( a-\frac{1}{2} \right )^{2}+\frac{3}{4}$

$x=2a-1,y=2b-1,z=2c-1$

$P=\frac{32(x+y+z-1)}{\prod (x^{2}+3)}$

Ta đi chứng minh : $\left ( y^{2}+3 \right )(z^{2}+3)\geq 4\left ( 1+\frac{\left ( y+z+1 \right )^{2}}{3} \right )$

$\Leftrightarrow 3(y-z)^{2}+3(yz-1)^{2}+2(y+z-2)^{2}\geq 0$

$\Rightarrow \prod (x^2+3)\geq 4\left ( x^2+3 \right )\left ( 1+\frac{(y+z+1)^{2}}{3} \right )\geq 4(x+y+z+1)^{2}$

Đặt : $t=x+y+z+1$

$t<2 \Rightarrow P<0$

$t\geq 2$

$0\leq P\leq \frac{32(t-2)}{4t^{2}}=\frac{8}{t}-\frac{16}{t^{2}}=-\left ( \frac{4}{t}-1 \right )^{2}+1\leq 1$

Dấu $"="\Leftrightarrow a=b=c=1$

$2+x+yz=x+2+\frac{yz}{2}+\frac{yz}{2}\geq 4\sqrt[4]{xyz}.\sqrt[4]{yz}.\sqrt[4]{\frac{1}{2}}$

$\Rightarrow \sum \frac{1}{2+x+yz}\leq \frac{1}{4\sqrt[4]{\frac{xyz}{2}}}\left ( \sum \frac{1}{\sqrt[4]{yz}} \right ) \leq \frac{1}{4\sqrt{2}}\left ( \sqrt{3\left ( \sum \frac{1}{\sqrt{yz}} \right )} \right ) \leq \frac{1}{4\sqrt{2}}\left ( \sqrt{3\sqrt{3\left ( \sum \frac{1}{yz} \right )}} \right )=\frac{3}{8}$