Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ap dông B§T C-S ta cã:
\(\frac{a}{a+\sqrt{2016a+bc}}=\frac{a}{a+\sqrt{\left(a+b+c\right)a+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\)
\(\le\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)
\(=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\). Tuong tù ta cx cã:
\(\frac{b}{b+\sqrt{2016b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}};\frac{c}{c+\sqrt{2016c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Céng theo vÕ c¸c B§T trªn ta dc:
\(VT\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
P/s:may mk bi loi Unikey r` mk dg ban chua kip chinh lai bn gang doc
Do \(a+b+c=1\) nên :
\(VT=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{ca}{b\left(a+b+c\right)+ac}}\)
\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)
Áp dụng BĐT AM - GM :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)
\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
Bài 2:
Chứng minh bất đẳng thức Mincopxki \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\text{ }\left(1\right)\)
(bình phương vài lần + biến đổi tương đương)
\(S\ge\sqrt{\left(a+b\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2}+\sqrt{c^2+\frac{1}{c^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{9}{a+b+c}\right)^2}\)
\(t=\left(a+b+c\right)^2\le\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(S\ge\sqrt{t+\frac{81}{t}}=\sqrt{t+\frac{81}{16t}+\frac{1215}{16t}}\ge\sqrt{2\sqrt{t.\frac{81}{16t}}+\frac{1215}{16.\frac{9}{4}}}=\frac{\sqrt{153}}{2}\)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{2}.\)
