Giải thích các bước giải:
Trước hết ta chứng minh bất đẳng thức :
$\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge \sqrt{(a+b)^2+(x+y)^2},a,b,x,y>0$
Thật vậy :
$\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge \sqrt{(a+b)^2+(x+y)^2}$
$\leftrightarrow (\sqrt{a^2+x^2}+\sqrt{b^2+y^2})^2\ge (\sqrt{(a+b)^2+(x+y)^2})^2$
$\leftrightarrow a^2+x^2+b^2+y^2+2\sqrt{a^2+x^2}.\sqrt{b^2+y^2}\ge (a+b)^2+(x+y)^2$
$\leftrightarrow a^2+x^2+b^2+y^2+2\sqrt{a^2+x^2}.\sqrt{b^2+y^2}\ge a^2+b^2+x^2+y^2+2(ab+xy)$
$\leftrightarrow \sqrt{a^2+x^2}.\sqrt{b^2+y^2}\ge(ab+xy)$
$\leftrightarrow \sqrt{(a^2+x^2)(b^2+y^2)}\ge ab+xy$ luôn đúng(BDT Bunhia)\to đpcm$
$\to \sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge \sqrt{(a+b)^2+(x+y)^2}+\sqrt{c^2+z^2}$
$\to \sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge \sqrt{(a+b+c)^2+(x+y+z)^2}$
$\to A=\sqrt{\dfrac{9}{(a+b)^2}+c^2}+\sqrt{\dfrac{9}{(b+c)^2}+a^2}+\sqrt{\dfrac{9}{(c+a)^2}+b^2}$
$\to A=\sqrt{(\dfrac{3}{a+b})^2+c^2}+\sqrt{(\dfrac{3}{b+c})^2+a^2}+\sqrt{(\dfrac{3}{c+a})^2+b^2}$
$\to A\ge\sqrt{(\dfrac{3}{a+b}+\dfrac{3}{b+c}+\dfrac{3}{c+a})^2+(a+b+c)^2}$
$\to A\ge\sqrt{9(\dfrac{9}{a+b+b+c+c+a})^2+(a+b+c)^2}$
$\to A\ge\sqrt{\dfrac{729}{4(a+b+c)^2}+(a+b+c)^2}$
$\to A\ge\sqrt{\dfrac{405}{4(a+b+c)^2}+\dfrac{81}{(a+b+c)^2}+(a+b+c)^2}$
$\to A\ge\sqrt{\dfrac{405}{4.3(a^2+b^2+c^2)}+2\sqrt{\dfrac{81}{(a+b+c)^2}.(a+b+c)^2}}$
$\to A\ge\sqrt{\dfrac{405}{4.3.3}+18}$
$\to A\ge \dfrac{3\sqrt{13}}{2}$
Dấu = xảy ra khi $a=b=c=1$
