Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta đi chứng minh BĐT phụ sau :
\(\sqrt[3]{4.\left(x^3+y^3\right)}\ge x+y\) với \(x,y>0\)
BĐT tương đương :
\(4.\left(a^3+b^3\right)\ge\left(a+b\right)^3\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)( đúng )
Áp dụng vào bài toán có :
\(P\ge a+b+b+c+c+a=2.\left(a+b+c\right)=2.2019=4038\)
Dấu "=" xảy ra khi \(a=b=c=673\)
Ta có: \(a+b+c+\sqrt{abc}=4\)
\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)
\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)
\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)
Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)
có cả mấy bất đẳng thức đó hả
bn viết công thức tổng quát ra cho mk vs
mk thanks
\(\frac{b\left(2a-b\right)}{a\left(b+c\right)}+\frac{c\left(2b-c\right)}{b\left(c+a\right)}+\frac{a\left(2c-a\right)}{c\left(a+b\right)}\le\frac{3}{2}\)
\(\Leftrightarrow\left[2-\frac{b\left(2a-b\right)}{a\left(b+c\right)}\right]+\left[2-\frac{c\left(2b-c\right)}{b\left(c+a\right)}\right]+\left[2-\frac{a\left(2c-a\right)}{c\left(a+b\right)}\right]\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\ge\frac{9}{2}\)
Áp dụng BĐT Schwarz, ta có :
\(\frac{b^2}{a\left(b+c\right)}+\frac{c^2}{b\left(c+a\right)}+\frac{a^2}{c\left(a+b\right)}\ge\frac{\left(a+b+c\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)( 1 )
\(\frac{ac}{a\left(b+c\right)}+\frac{ab}{b\left(c+a\right)}+\frac{bc}{c\left(a+b\right)}=\frac{c^2}{c\left(b+c\right)}+\frac{a^2}{a\left(a+c\right)}+\frac{b^2}{b\left(a+b\right)}\) ( 2 )
\(\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ac}\)
Cộng ( 1 ) với ( 2 ), ta được :
\(\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\)
\(\ge\left(a+b+c\right)^2\left(\frac{1}{2\left(ab+bc+ac\right)}+\frac{2}{a^2+b^2+c^2+ab+bc+ac}\right)\)
\(\ge\left(a+b+c\right)^2\left(\frac{\left(1+2\right)^2}{2\left(ab+bc+ac\right)+2\left(a^2+b^2+c^2+ab+bc+ac\right)}\right)=\frac{9}{2}\)
không biết cách này ổn không
Ta có : \(\frac{b\left(2a-b\right)}{a\left(b+c\right)}=\frac{2-\frac{b}{a}}{\frac{c}{b}+1}\) ; tương tự :...
đặt \(\frac{a}{c}=x;\frac{b}{a}=y;\frac{c}{b}=z\Rightarrow xyz=1\)
\(\Sigma\frac{2-y}{z+1}\le\frac{3}{2}\)
\(\Leftrightarrow2\Sigma xy^2+2\Sigma x^2+\Sigma xy\ge3\Sigma x+6\)( quy đồng khử mẫu )
\(\Leftrightarrow\Sigma\frac{x}{y}\ge\Sigma x\)( xyz = 1 ) ( luôn đúng )
\(\Rightarrowđpcm\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+2c\right)}=\frac{\sqrt{3a\left(b+2c\right)}}{\sqrt{3}}\le\frac{\frac{3a+b+2c}{2}}{\sqrt{3}}=\frac{3a+b+2c}{2\sqrt{3}}\)
Tương tự ta cũng có:\(\sqrt{b\left(c+2a\right)}\le\frac{3b+c+2a}{2\sqrt{3}}\)
\(\sqrt{c\left(a+2b\right)}\le\frac{3c+a+2b}{2\sqrt{3}}\)
Cộng theo vế các BĐT lại ta được:
\(VT\le\frac{3a+b+2c}{2\sqrt{3}}+\frac{3b+c+2a}{2\sqrt{3}}+\frac{3c+a+2b}{2\sqrt{3}}=\frac{6a+6b+6c}{2\sqrt{3}}=\frac{6.4}{2\sqrt{3}}=4\sqrt{3}\)
Câu 3:
bạn cứ áp dụng cái \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Câu 4:
từ giả thiết :\(a+b+c+\sqrt{abc}=4\Leftrightarrow\sqrt{abc}=4-a-b-c\Leftrightarrow abc=\left(4-a-b-c\right)^2\)
ta có: \(a\left(4-b\right)\left(4-c\right)=a\left(16-4c-4b+bc\right)=16a-4ac-4ab+abc\)
\(=16a-4ab-4ac+\left[4-\left(a+b+c\right)\right]^2=16a-4ab-4ac+16-8\left(a+b+c\right)+\left(a+b+c\right)^2\)
\(=a^2+b^2+c^2-2ab-2ac+2bc+8a-8b-8c+16\)
\(=\left(a-b-c\right)^2+8\left(a-b-c\right)+16=\left(a-b-c+4\right)^2\)
\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=a-b-c+4\)(vì \(a-b-c+4=a-b-c+a+b+c+\sqrt{abc}=2a+\sqrt{abc}>0\))
các căn thức còn lại tương tự ...
a;b;c có dương ko bạn?
Ko dương thì biểu thức này ko tồn tại min đâu