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Dat $t=\sqrt{x^2+1}-x\Rightarrow \sqrt{x^2+1}+x=\frac{1}{t},$ PT da cho tro thanh:
$t^5+\frac{1}{t^5}=123\Leftrightarrow t^{10}-123t^5+1=0$
Toi day thi a nghi e lam duoc roi..
Ap dung bat dang thuc $(ab+cd)^2 \le (a^2+c^2)(b^2+d^2),$ ta co:
$(\sqrt{x-1}+\sqrt{y-2})^2 \le (1^2+1^2)(x-1+y-2)=2$ (vi $x+y=4$)
$\Rightarrow \sqrt{x-1}+\sqrt{y-2} \le \sqrt{2}.$
GTLN dat duoc tai $x=\frac{3}{2};y=\frac{5}{2}.$