Trích:
Nguyên văn bởi thinh tran  Cho các số thực $a;\,b;\,c$ khác 0, chứng minh rằng
\[\sqrt[4]{{\frac{{{a^2}}}{{{a^2} - ab + {b^2}}}}} + \sqrt[4]{{\frac{{{b^2}}}{{{b^2} - bc + {c^2}}}}} + \sqrt[4]{{\frac{{{c^2}}}{{{c^2} - ca + {a^2}}}}} \le 3.\] |
Ta có:
$$a^2-ab+b^2=\frac{1}{4}\left ( a+b \right )^{2}+\frac{3}{4}(a-b)^{2}\geq \frac{1}{4}(a+b)^2$$
$$\Rightarrow \sqrt[4]{\frac{a^2}{a^2-ab+b^2}}\leq \sqrt{2}\sqrt{\frac{a}{a+b}}$$
Ta cần chứng minh:
$$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{ \frac{c}{c+a}}\leq \frac{3}{\sqrt{2}}$$
Đây là bất đẳng thức quen thuộc
$$VT^2=\left ( \sqrt{\frac{a(b+c)}{(a+b)(b+c)}}+\sqrt{\frac{b(c+a )}{(b+c)(c+a)}}+\sqrt{\frac{c(a+b)}{(c+a)(a+b)}} \right )^{2}$$
$$<=\left [ a(b+c)+b(c+a)+c(a+b) \right ]\left ( \frac{1}{(a+b)(b+c)} +\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}\right )= $$$$4\frac{(ab+bc+ca)(a+b+c)}{(a+b)(b+c)(c+a)}=$$
$$4\left [ \frac{9}{8}-\frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{8(a+b)(b+c)(c+a)} \right ]\leq 4.\frac{9}{8}=\frac{9}{2}
\Rightarrow VT\leq \frac{3}{\sqrt{2}}$$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]