Cho a,b,c >0, a+b<= c; Tìm GTNN của R= (a^2+b^2+c^2) * (1/a^2+1/b^2+1/c^2)
[tex]R=(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})=3+\frac{a^2+b^2}{c^2}+c^2(\frac{1}{a^2}+\frac{1}{b^2})+\frac{a^2}{b^2}+\frac{b^2}{a^2}\\=3+(\frac{a^2}{c^2}+\frac{c^2}{16a^2})+(\frac{b^2}{c^2}+\frac{c^2}{16b^2})+\frac{15c^2}{16}(\frac{1}{a^{2}}+\frac{1}{b^2})+(\frac{x^2}{y^2}+\frac{y^2}{x^2})[/tex]
Áp dụng bđt Cauchy cho 4 ngoặc ta có
[tex]R\geq 3+\frac{1}{2}+\frac{1}{2}+\frac{15c^2}{16}.\frac{2}{ab}+2\geq 6+\frac{15c^2}{16}.\frac{2}{(\frac{a+b}{2})^{2}}\doteq 6+\frac{15c^2}{16}.\frac{8}{(a+b)^2}\\\geq 6+\frac{15}{2}.(\frac{c}{a+b})^{2}\geq 6+\frac{15}{2}=\frac{27}{2}[/tex]
Dấu = xảy ra khi [tex]a=b=\frac{c}{2}[/tex]
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