vutienhoang

P=$\frac{2+a}{\sqrt{1+b}}+\frac{2+b}{\sqrt{1+a}}$=$\frac{1}{3}(\frac{1+a}{\sqrt{1+b}}+\frac{1+b}{\sqrt{1+a}})+\frac{2}{3}\frac{1+a}{\sqrt{1+b}}+\frac{2(1+b)}{3\sqrt{1+a}}+\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}\geq \frac{1}{3}(\frac{(2+a+b)^2}{\sqrt{(1+a)(1+b)}(\sqrt{1+a}+\sqrt{1+b})})+4\sqrt{\frac{2}{3}}\geq \frac{3}{(\frac{a+b+2}{2})(\sqrt{1+a}+\sqrt{1+b})}+4\sqrt{\frac{2}{3}}$
có$\sqrt{\frac{3}{2}(1+a)}+\sqrt{\frac{3}{2}(1+b)}\leq \frac{\frac{5}{2}+a+\frac{5}{2}+b}{2}=3 =>\sqrt{1+a}+\sqrt{1+b}\leq 3\sqrt{\frac{2}{3}}$

 đặt A =a+b+c-abc
$A^{2}\leq (a^2+1)((bc-1)^2+(b+c)^2)=(a^2+1)(b^2+1)(c^2+1)\leq (\frac{a^2+b^2+c^2+3}{3})^3=8$
<=>$A\leq 2\sqrt{2}$
 

$\sum \frac{a^3}{b+c}=\sum \frac{a^4}{ab+bc}\geq \frac{(\sum a^2)^2}{2\sum ab}\geq \frac{\sum a^2}{2}$
=> $3\sum \frac{a^3}{b+c}\geq \sum \frac{a^3}{b+c}+\sum a^2=\sum \frac{a^2(a+b+c)}{b+c}$
=> ta cần cm $\sum \frac{a^2}{b+c}\geq \sum \frac{a}{b+c}. (\frac{a+b+c}{3})=\sum \frac{a^2}{3(b+c)}+\sum \frac{b+c}{3} <=> \frac{2a^2}{3(b+c)}\geq \frac{2(a+b+c)}{3}$  ( đúng theo cauchy-schwarz)

bđt <=>$(a^2-1)(2a^2-1)+\frac{1}{a^2+1}\geq 0$
nếu $a^2\geq$1 hoăc $a^2\leq 1/2$ thì bđt luôn đúng
nếu $1\geq a^2\geq \frac{1}{2}$ có $(a^2-1)(2a^2-1)=-(1-a^2)(2a^2-1)\geq -(\frac{a^2}{2})^2\geq -\frac{1}{4}$ có $\frac{1}{a^2+1}\geq \frac{1}{2}$ => bđt đúng

nhân pt 2 với số ảo i cộng với pt 1 có$x+yi+\frac{3x-y-xi-3yi}{x^2+y^2}=3$
đặt z =x+yi =>$\frac{1}{z}=\frac{x-yi}{x^2+y^2}$
=> $z+\frac{3-i}{z}=0 <=>z^2-3z+3-i=0 <=>z=2+i$ hoặc z=1-i
th1 z=2+i => x+yi =2+i => x=2 y=1
th2 z=1-i => x+yi=1-i => x=1 y=-1
thử lại thỏa mãn

$(\sum \sqrt{a-1})^2\leq (1+c-1)((\sqrt{a-1}+\sqrt{b-1})^2+1)=c(a+b-1+2\sqrt{(a-1)(b-1)})=c(ab+1-(ab-a-b+1-2\sqrt{(a-1)(b-1)}+1))=c(ab+1-(\sqrt{(a-1)(b-1)}-1)^2)\leq c(ab+1)$

pt 1 <=>$2(x-1)= (y-1)(y^2+2y+3)$
tương tự với 2 pt còn lại nhân 2 vế cảu 3 pt => $8(x-1)(y-1)(z-1)=(y-1)(z-1)(x-1)(x^2+2x+3)(y^2+2y+3)(z^2+2z+3)$
<=>th1 x-1=0 =>x=1 thay vào => y =1 và z=1
th2 y-1 = 0 tương tự => x=z=1
th3 z-1=0 =>x=y=1
th4 8=(x^2+2x+3)(y^2+2y+3)(z^2+2z+3)$
có $x^2+2x+3=(x+1)^2+2$$\geq$2 =>(x^2+2x+3)(y^2+2y+3)(z^2+2z+3)$ $\geq$ 8 dẫu = xảy ra  <=> x=y=z =-1
thử lại có x=y=z=-1 hoặc x=y=z =1

2) pt 1 <=> $\sqrt[4]{a-1}=\sqrt{b-1}+1\geq 1$ => a$\geq$ 2 = > $a^{2}\geq 4$ có  $b \geq 1$ => $a^{2}+2b\geq 6$

có  1-2$cos\alpha +cos\beta +cos\gamma =1-sin^{2}\ \frac{\alpha }{2}+2cos\frac{\beta +\gamma }{2}cos\frac{\beta -\gamma }{2}$
=$1-2sin\frac{\alpha }{2}cos\frac{\beta +\gamma }{2}+2sin\frac{\alpha }{2}cos\frac{\beta -\gamma }{2}=1-2sin\frac{\alpha }{2}(cos\frac{\beta +\gamma }{2}-cos\frac{\beta -\gamma }{2})$ =1 + 4$sin\frac{\alpha }{2}sin\frac{\beta }{2}sin\frac{\gamma }{2}$

đk $x^{2}-\sqrt{3}\geq 0$
pt,<=>$(x^2-\sqrt{3})(x+\sqrt{x^2-\sqrt{3}})+(x^2+\sqrt{3})(\sqrt{x^2+\sqrt{3}}-x)= \sqrt{3}x$
<=>$\sqrt{x^2-\sqrt{3}}^3+\sqrt{x^2+\sqrt{3}}^3=3\sqrt{3}x$
<=>$(x^2-\sqrt{3})^3+2\sqrt{x^4-3}^3+(x^2+\sqrt{3})^3= 27x^2$
<=>$2\sqrt{x^4-3}^3= 9x^2-2x^6$
=>$4(x^4-3)^3=(9x^2-2x^6)^2$
<=>$27x^4-108=0$

$(\sum \frac{a}{\sqrt{b^2+8}})^2(\sum a(b^2+8))\geq (a+b+c)^3$  (bđt holer)
từ đk => a+b+c= 3abc nên$\sum a(b^{2}+8) = \sum ab^{2} +8\sum a= \sum ab^{2} +24abc$ .
$(a+b+c)^{3}= \sum a^{3}+3\sum ab^2 +3\sum a^2b +6abc \geq \sum ab^2 +24abc$   nên$\sum \frac{a}{\sqrt{b^{2}+8}}$$\geq$ 1