cho a, b, c > 0 thỏa mãn abc = 1. Cmr: \(\frac{1}{\sqrt{a^5-a^2+3ab+6}}+\frac{1}{\sqrt{b^5-b^2+3bc}+6}+\frac{1}{\sqrt{c^5-c^2+3ca+6}}\le1\)
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30 tháng 11 2019
Ta có: \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}=\frac{a^2+ab+1}{\sqrt{a^2+ab+2ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+a^2+b^2+c^2}}=\sqrt{a^2+ab+1}\)
\(\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}=\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2}\)
\(=\frac{1}{\sqrt{5}}.\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2\right)}\)
\(\ge\frac{1}{\sqrt{5}}\sqrt{\left(\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{2}b+a+c\right)^2}\)
\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
Tương tự ta cũng chứng minh đc:
\(\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}b+\frac{3}{2}c+a\right)\)
\(\frac{c^2+ca+1}{\sqrt{c^2+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}c+\frac{3}{2}a+b\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}+\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}+\frac{c^2+ca+1}{\sqrt{c^3+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(5a+5b+5c\right)\)
\(=\sqrt{5}\left(a+b+c\right)\)
Dấu "=" xảy ra <=> a = b = c =\(\frac{1}{\sqrt{3}}\)
TN
30 tháng 10 2016
Ta có:\(a^5+ab+b^2\ge3a^2b\)
Tương tự ta có:
\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)
\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)
Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)
\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)
Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)
\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
Giả sử \(xy\le1\) thì \(z\ge1\)
Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)
\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)
Dấu = khi \(a=b=c=1\)
1 tháng 1 2020
Áp dụng liên tiếp AM - GM và Cauchy - Schwarz ta có :
\(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left(a^2+b^2\right)}}\)
\(=\frac{a^2+ab+1}{\sqrt{a^2+ab+1}}\)
\(=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\)
\(=\frac{1}{\sqrt{5}}\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+a^2+c^2\right]}\)
\(\ge\frac{1}{\sqrt{5}}\left[\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{4}b+a+c\right]\)
\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
Chứng minh tương tự và công lại ta có đpcm
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
HF
24 tháng 10 2019
\(\sqrt{a^2+3a+5}\ge\frac{5a+13}{6}\Leftrightarrow a^2+3a+5\ge\frac{25a^2+130a+169}{36}\)
\(\Leftrightarrow36a^2+108a+180\ge25a^2+130a+169\Leftrightarrow11a^2-22a+11\ge0\)
\(\Leftrightarrow11\left(a-1\right)^2\ge0\forall a\inℝ\)
Dấu = xảy ra khi a=1
24 tháng 10 2019
Ta có:
\(\sqrt{a^2+3ab+5b^2}=\sqrt{\left(\frac{25a^2}{36}+\frac{130ab}{36}+\frac{169}{36}\right)+\frac{11}{36}\left(a^2-2ab+b^2\right)}\)
\(=\sqrt{\left(\frac{5a}{6}+\frac{13b}{6}\right)^2+\frac{11}{36}\left(a-b\right)^2}\ge\frac{5a+13b}{6}\)
Tương tự:\(\sqrt{b^2+3bc+5c^2}\ge\frac{5b+13c}{6};\sqrt{c^2+3ca+5a^2}\ge\frac{5c+13a}{6}\)
Khi đó:\(P=\sqrt{a^2+3ab+5b^2}+\sqrt{b^2+3bc+5c^2}+\sqrt{c^2+3ac+5a^2}\)
\(\ge\frac{5a+13b+5b+13c+5c+13a}{6}=\frac{18\left(a+b+c\right)}{6}=3\left(a+b+c\right)=9\)
Dấu "=" xảy ra tại \(a=b=c=1\)