Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
làm như giỏi lắm í, thôi khỏi nói cũng biết, ko cần thể hiện đâu
\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a+b+c}{\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}}\)
Ta có: \(\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}\)
\(=\sqrt{ab+bc+ac+a^2}+\sqrt{ab+bc+ac+b^2}+\sqrt{ab+bc+ca+c^2}\)
\(=\sqrt{b\left(a+c\right)+a\left(a+c\right)}+\sqrt{b\left(a+b\right)+c\left(a+b\right)}+\sqrt{b\left(a+c\right)+c\left(a+c\right)}\)
\(=\sqrt{\left(a+c\right)\left(a+b\right)}+\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(\le\frac{a+c+a+b}{2}+\frac{a+b+b+c}{2}+\frac{a+c+b+c}{2}\)
\(\le\frac{2a+a+2b+b+2c+c}{2}=\frac{3a+3b+3c}{2}=\frac{3}{2}\left(a+b+c\right)\)
Suy ra : \(A=\frac{a+b+c}{\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}}\ge\frac{2}{3}\)
Dấu "=" xảy ra khi và chỉ khi a=b=c=0
Vậy Amin = \(\frac{2}{3}\)
Chắc sai. Mong bạn giúp đỡ. Cảm ơn!
Sửa lại đề là tìm Max nhé m.n
Ta có:
\(\frac{ab+bc+ca+6\left(a+b+c\right)+27}{\left(a+3\right)\left(b+3\right)\left(c+3\right)}=\frac{3}{5}\)
\(\Leftrightarrow\frac{\left(b+3\right)\left(c+3\right)+\left(c+3\right)\left(a+3\right)+\left(a+3\right)\left(b+3\right)}{\left(a+3\right)\left(b+3\right)\left(c+3\right)}=\frac{3}{5}\)
\(\Leftrightarrow\frac{5}{a+3}+\frac{5}{b+3}+\frac{5}{c+3}=3\Leftrightarrow\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}=0\)
Xét biểu thức:
\(\frac{a^2-4}{a^2-9}=\frac{\left(a-2\right)\left(a+2\right)}{\left(a-3\right)\left(a+3\right)}=\frac{a-2}{a+3}.\frac{a+2}{a-3}\)
tưởng tự:
\(\frac{b^2-4}{b^2-9}=\frac{b-2}{b+3}.\frac{b+2}{b-3},\frac{c^2-4}{c^2-9}=\frac{c-2}{c+3}.\frac{c+2}{c-3}\)
\(\Rightarrow\frac{a^2-4}{a^2-9}+\frac{b^2-4}{b^2-9}+\frac{c^2-4}{c^2-9}=\frac{a-2}{a+3}.\frac{a+2}{a-3}+\frac{b-2}{b+3}.\frac{b+2}{b-3}+\frac{c-2}{c+3}.\frac{c+2}{c-3}\)
Do vai trò của a và b và c như nhau nên ta giả sử
\(a\ge b\ge c\)
Khi đó ta có:
\(\frac{a-2}{a+3}\ge\frac{b-2}{b+3}\ge\frac{c-2}{c+3},\frac{a+2}{a-3}\le\frac{b+2}{b-3}\le\frac{c+2}{c-3}\)
Áp dụng bất đẳng thức chebyshev cho 2 bộ ngược chiều trên ta có
\(\frac{a-2}{a+3}.\frac{a+3}{a-2}+\frac{b-2}{b+3}.\frac{b+2}{b-3}+\frac{c-2}{c+3}.\frac{c+2}{c-3}\le\left(\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}\right).\left(\frac{a+2}{a-3}+\frac{b+2}{b-3}+\frac{c+2}{c-3}\right)\)
Mà \(\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}=0\)
\(\Rightarrow\frac{a^2-4}{a^2-9}+\frac{b^2-4}{b^2-9}+\frac{c^2-4}{c^2-9}\le0\)
\(\Rightarrow\frac{5}{a^2-9}+\frac{5}{b^2-9}+\frac{5}{c^2-9}\le-3\Rightarrow\frac{1}{a^2-9}+\frac{1}{b^2-9}+\frac{1}{c^2-9}\le\frac{-3}{5}\)
Dấu bằng xảy ra khi a=b=c=2
Đúng như bạn Quang viết, GTNN của S là 13 khi \(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\), nhưng mình cần một lời giải thích vì sao nó lại ra như vậy.
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
Cho a,b,c>0 và a+b+c=1. CMR: \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{3}{2}\)
Ta có : a + bc = a ( a + b + c ) + bc = ( a + c ) ( a + b )
BĐT cần chứng minh tương đương với :
\(\frac{a\left(a+b+c\right)-bc}{\left(a+c\right)\left(a+b\right)}+\frac{b\left(a+b+c\right)-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c\left(a+b+c\right)-ab}{\left(c+a\right)\left(c+b\right)}\le\frac{3}{2}\)
\(\left(a^2+ab+ac-bc\right)\left(b+c\right)+\left(ab+b^2+bc-ac\right)\left(a+c\right)+\left(ac+bc+c^2-ab\right)\left(a+b\right)\le\frac{3}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
khai triển ra , ta được :
\(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+6abc\le\frac{3}{2}\left(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\right)+3abc\)
\(\Rightarrow\frac{-1}{2}\left(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\right)\le-3abc\)
\(\Rightarrow a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\ge6abc\)( nhân với -2 thì đổi dấu )
\(\Rightarrow b\left(a^2-2ac+c^2\right)+a\left(b^2-2bc+c^2\right)+c\left(a^2-2ab+b^2\right)\ge0\)
\(\Rightarrow b\left(a-c\right)^2+a\left(b-c\right)^2+c\left(a-b\right)^2\ge0\)
vì BĐT cuối luôn đúng nên BĐT lúc đầu đúng
Dấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
bạn tham khảo nhé : https://olm.vn/hoi-dap/detail/222370673956.html
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
quy đồng ,bdt cần cm <=> (4-a)(4-b)(4-c) >= 27abc
<=>ab+bc+ca >= 3abc
amgm VT ,dpcm <=> 3.căn bậc 3((abc)2) >/ 3abc <=> abc <= 1
4=(a+b+c)+abc >/ 3.căn bậc 3(abc)+abc , giải bpt -> abc <= 1