Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(a^2+b^2+c^2\ge ab+ac+\)\(bc\)(1)
vì , ta có
(1) \(\Leftrightarrow\)\(2\left(a^2+b^2+c^2\right)\)\(\ge2\left(ab+ac+bc\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)\)\(+\left(a^2-2ac+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng) => bất đẳng thức
Ta có :
\(a^2+b^2+c^2-2abc\ge ab+bc+ac-2abc\)
<=>\(a^2+b^2+c^2+2abc-3abc\ge ab+bc+ac-2abc\)
<=> \(1-3abc\ge ab+bc+ac-2abc\)
=> MAX P=1 <=> \(\hept{\begin{cases}a=0\\b=c=1\end{cases}}\)hoặc \(\hept{\begin{cases}b=0\\a=c=1\end{cases}}\)
hoặc \(\hept{\begin{cases}c=0\\a=b=1\end{cases}}\)
Sai thì bảo mình nhé
xin lỗi Dòng thứ 8 và 9 phải là
\(a^2+b^2+c^2+2abc-4abc\ge ab+ac+bc-2abc\)
\(\Leftrightarrow1-4abc\ge ab+ac+bc-2abc\)
ta có: a,b,c>0 mà a+b+c=1 \(\Rightarrow\left(1-a\right)\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a\left(a-b\right)^2\le\left(a-b\right)^2\)
tương tự và cộng theo vế: \(VT\le6\left(ab+bc+ca\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=2\left(a+b+c\right)^2=2\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Câu hỏi của nguyen thu phuong - Toán lớp 8 - Học toán với OnlineMath
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
$\text{VT}=\frac{a(a+b+c)+bc}{b+c}+\frac{b(a+b+c)+ac}{a+c}+\frac{c(a+b+c)+ab}{a+b}$
$=\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}+\frac{(c+a)(c+b)}{a+b}$
Áp dụng BĐT AM-GM:
$\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}\geq 2\sqrt{(a+b)^2}=2(a+b)$
$\frac{(b+c)(b+a)}{a+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(b+c)^2}=2(b+c)$
$\frac{(a+b)(a+c)}{b+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(c+a)^2}=2(a+c)$
Cộng các BĐT trên theo vế và thu gọn:
$\text{VT}\geq 2(a+b+c)=2$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
\(ab+bc+ca=abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
\(\frac{a}{bc\left(a+1\right)}=\frac{\frac{1}{x}}{\frac{1}{y}\cdot\frac{1}{z}\left(\frac{1}{x}+1\right)}=\frac{xyz}{x\left(x+1\right)}=\frac{yz}{x+1}\)
Tươn tự rồi cộng vế theo vế:
\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{\left(x+y\right)^2}{4\left(z+1\right)}+\frac{\left(y+z\right)^2}{4\left(x+1\right)}+\frac{\left(z+x\right)^2}{4\left(y+1\right)}\)
Đặt \(x+y=p;y+z=q;z+x=r\Rightarrow p+q+r=2\)
\(A\le\Sigma\frac{\left(x+y\right)^2}{4\left(z+1\right)}=\Sigma\frac{\left(x+y\right)^2}{4\left[\left(z+y\right)+\left(z+x\right)\right]}=\frac{p^2}{4\left(q+r\right)}+\frac{r^2}{4\left(p+q\right)}+\frac{q^2}{4\left(p+r\right)}\)
Sau khi đổi biến,cô si thì em ra thế này.Ai đó giúp em với :)
Áp dụng Côsi:
\(2.\frac{4}{3}.\sqrt{2a+bc}\le\left(\frac{4}{3}\right)^2+2a+bc\)
Tương tự: \(2.\frac{4}{3}\sqrt{2b+ca}\le\frac{16}{9}+2b+ca;2.\frac{4}{3}\sqrt{2c+ab}\le\frac{16}{9}+2c+ab\)
\(\Rightarrow\frac{8}{3}Q\le\frac{16}{3}+2\left(a+b+c\right)+bc+ca+ab=\frac{28}{3}+ab+bc+ca\)
Ta có: \(3\left(ab+bc+ca\right)=2\left(ab+bc+ca\right)+ab+bc+ca\)
\(\le2\left(ab+bc+ca\right)+a^2+b^2+c^2=\left(a+b+c\right)^2=4\)
\(\Rightarrow ab+bc+ca\le\frac{4}{3}\)
\(\Rightarrow\frac{8}{3}Q\le\frac{28}{3}+\frac{4}{3}=\frac{32}{3}\Rightarrow Q\le4\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{2}{3}\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
P=\(\left(a^2+b^2+c^2+2ab+2ac+2bc\right)+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\)\(+a^3+b^3+c^3-2\left(a^2b+b^2c+c^2a\right)+ab^2+bc^2+ca^2\)\(=1+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\left(a+b+c\right)+\left(a^3+b^3+c^3\right)\)\(-2\left(a^2b+b^2c+c^2a\right)+\left(ab^2+bc^2+ca^2\right)\)\(=1+4\left(ab+bc+ca\right)-3\left(a^2b+b^2c+c^2a\right)\)
Mà \(\left(a^2b+b^2c+c^2a\right)\left(b+c+a\right)\ge\left(ab+bc+ca\right)^2\)
=> \(P\le1+4\left(ab+bc+ca\right)-3\left(ab+bc+ca\right)^2\). Đặt \(ab+bc+ca=t\le\frac{1}{3}\)
=> \(P\le-3\left(t^2-\frac{2}{3}t+\frac{1}{9}\right)+2t+\frac{4}{3}\le-3\left(t-\frac{1}{3}\right)^2+\frac{2}{3}+\frac{4}{3}\le2\)
Dấu bằng xảy ra khi \(t=\frac{1}{3}\)<=> \(a=b=c=\frac{1}{3}\)