Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{ax+by+cz}{xy}=z\Rightarrow z=\frac{a}{y}+\frac{b}{x}+\frac{cz}{xy}>\frac{a}{y}+\frac{b}{x}\)
Tương tự có \(y>\frac{a}{z}+\frac{c}{x}\); \(x>\frac{b}{z}+\frac{c}{y}\)
\(\Rightarrow x+y+z>\frac{b+c}{x}+\frac{a+c}{y}+\frac{a+b}{z}=\frac{b+c}{x}+x+\frac{a+c}{y}+y+\frac{a+b}{z}+z-x-y-z\)
\(\Rightarrow2\left(x+y+z\right)>2\sqrt{b+c}+2\sqrt{a+c}+2\sqrt{a+b}\)
\(\Rightarrow x+y+z>\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đầu tiên chứng minh:
\(\left(a^2x+b^2y+c^2z\right)\left(yz+zx+xy\right)\ge xyz\left(a+b+c\right)^2\)
\(=xyz\left(x+z+y\right)^2\ge3xyz\left(xy+yz+zx\right)\)
\(\Rightarrow a^2x+b^2y+c^2z\ge3xyz\)
Tương tự có:
\(x^2a+y^2b+z^2c\ge3abc\)
\(\Rightarrow\) ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1b/
Áp dụng BĐT Cô-si :
\(\sqrt{\frac{b+c}{a}}\le\frac{\frac{b+c}{a}+1}{2}=\frac{\frac{a+b+c}{a}}{2}=\frac{a+b+c}{2a}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)
Chứng minh tương tự:
\(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\); \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
Cộng theo vế ta được :
\(VT\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu "=" không xảy ra nên \(VT>2\).
2a/ Chắc là tính GT của \(x+y\).
\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\)
\(\Leftrightarrow\left(x-\sqrt{x^2+2013}\right)\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\)
Do vai trò \(x,y\) là như nhau nên thiết lập tương tự ta có :
\(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\)
Cộng theo vế 2 pt ta được :
\(x+y+\sqrt{x^2+2013}+\sqrt{y^2+2013}=\sqrt{x^2+2013}+\sqrt{y^2+2013}-x-y\)
\(\Leftrightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
Vậy....
2b/
Đặt \(A=5a^2+15ab-b^2\) và \(B=3a+b\)
Ta có \(B^2=\left(3a+b\right)^2=9a^2+6ab+b^2\)
Lấy \(A+B^2=5a^2+15a-b^2+9a^2+6ab+b^2\)
\(A+B^2=14a^2+21ab\)
\(A+B^2=7\left(2a+3ab\right)⋮7\)
Mà \(A⋮7\) ( vì \(A⋮49\) ) nên \(B^2⋮7\)
Vì 7 nguyên tố nên \(B⋮7\) ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này hình như có lần làm rồi :))
Đặt `ax^3=by^3=cz^3=k^3`
`=>a=k^3/x^3,b=k^3/y^3,c=k^3/z^3`
`=>root{3}{a}+root{3}{b}+root{3}{c}=k/x+k/y+k/z=k(1/x+1/y+1/z)=k(1)`
`**:ax^2+by^2+cz^2=(ax^3)/x+(by^3)/y+(cz^3)/z=k^3/x+k^3/y+k^3/z=k^3(1/x+1/y+1/z)=k^3`
`=>root{3}{ax^2+by^2+cz^2}=k(2)`
`(1)(2)=>ĐPCM`
Ta có :
\(x=\frac{ax}{yz}+\frac{b}{z}+\frac{c}{y}\)
\(y=\frac{a}{z}+\frac{by}{zx}+\frac{c}{x}\)
\(z=\frac{a}{y}+\frac{b}{x}+\frac{xy}{cz}\)
\(\Rightarrow\)\(x+y+z=\left(\frac{ax}{yz}+\frac{by}{zx}+\frac{cz}{xy}\right)+\frac{b+c}{x}+\frac{c+a}{y}+\frac{a+b}{z}>\frac{b+c}{z}+\frac{c+a}{y}+\frac{a+b}{z}\)
\(\ge\frac{\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2}{x+y+z}\)
\(\Leftrightarrow\)\(\left(x+y+z\right)^2>\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
\(\Leftrightarrow\)\(x+y+z>\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\) ( đpcm )