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vì có 1 chút nhầm lẫn nên giờ mk mới ra mong bạn thứ lỗi
bài 1
\(\Leftrightarrow\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2c^2b^2}+\frac{4c^4}{2c^3+2a^2c^2}\)
\(\ge\frac{\left(2a^2+2b^2+2c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\)
Dấu bằng xảy ra khi \(a=b=c=1\)
Bài 2 là chuyên Bình Thuận, 2016-2017
Áp dụng bất đẳng thức Cauchy – Schwarz, ta có:
\(\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Tương tự: \(\frac{yz}{y^2+zx+xy}\le\frac{xy\left(z^2+zx+xy\right)}{\left(xy+yz+zx\right)^2}\);\(\frac{zx}{z^2+xy+yz}\le\frac{zx\left(x^2+xy+yz\right)}{\left(xy+yz+zx\right)^2}\)
Cộng từng vế của 3 BĐT trên. ta được:
\(VT\le\frac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{\left(xy+yz+zx\right)^2}=\frac{x^2+y^2+z^2}{xy+yz+zx}\)
Đẳng thức xảy ra khi x = y = z
Đặt \(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}=k\)
\(\Rightarrow\begin{cases}a=\frac{x^2-yz}{k}\\b=\frac{y^2-zx}{k}\\c=\frac{z^2-xy}{k}\end{cases}\)
Ta có:
\(\frac{a^2-bc}{x}=\frac{\left(\frac{x^2-yz}{k}\right)^2-\frac{y^2-zx}{k}.\frac{z^2-xy}{k}}{x}=\frac{\frac{x^4-2x^2yz+\left(yz\right)^2}{k^2}-\frac{\left(y^2-zx\right).\left(z^2-xy\right)}{k^2}}{x}\)
\(=\frac{\frac{\left(x^4-2x^2yz+y^2z^2\right)-\left(y^2z^2-z^3x-xy^3+x^2zy\right)}{k^2}}{x}\)
\(=\frac{\frac{x^4-2x^2yz+y^2z^2-y^2z^2+z^3x+xy^3-x^2zy}{k^2}}{x}=\frac{x^4++z^3x+xy^3-3x^2yz}{k^2}.\frac{1}{x}=\frac{x^3+y^3+z^3-3xyz}{k^2}\)
Tương tự thay a;b;c vào \(\frac{b^2-ca}{y};\frac{c^2-ab}{z}\) ta cũng được \(\frac{b^2-ca}{y}=\frac{c^2-ab}{z}=\frac{x^3+y^3+z^3-3xyz}{k^2}\)
Vậy \(\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\left(đpcm\right)\)
\(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\)
\(\Leftrightarrow\)\(\frac{a}{x^2-yz}=\frac{b}{y^2-zx}=\frac{c}{z^2-xy}\)
\(\Leftrightarrow\)\(\frac{a^2}{\left(x^2-yz\right)^2}=\frac{b^2}{\left(y^2-zx\right)^2}=\frac{c^2}{\left(z^2-xy\right)^2}=\frac{ab}{\left(x^2-yz\right)\left(y^2-zx\right)}=\frac{bc}{\left(y^2-zx\right)\left(z^2-xy\right)}=\frac{ca}{\left(z^2-xy\right)\left(x^2-yz\right)}\left(1\right)\)
Áp dụng tính chất tỉ lệ thức ta có:
\(\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-zx\right)\left(z^2-xy\right)}=\frac{a^2-bc}{\left(x^2-yz\right)^2-\left(y^2-zx\right)\left(z^2-xy\right)}=\frac{a^2-bc}{x\left(x^3+y^3+z^3-3xyz\right)}\) (2)
\(\frac{b^2}{\left(y^2-zx\right)^2}=\frac{ac}{\left(x^2-yz\right)\left(z^2-xy\right)}=\frac{b^2-ac}{\left(y^2-zx\right)^2-\left(x^2-yz\right)\left(z^2-xy\right)}=\frac{b^2-ca}{y\left(x^3+y^3+z^3-3xyz\right)}\) (3)
\(\frac{c^2}{\left(z^2-xy\right)}=\frac{ab}{\left(x^2-yz\right)\left(y^2-xz\right)}=\frac{c^2-ab}{\left(z^2-xy\right)-\left(x^2-yz\right)\left(y^2-xz\right)}=\frac{c^2-ab}{z\left(x^3+y^3+z^3-3xyz\right)}\) (4)
Từ (1), (2), (3), (4) suy ra:
\(\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\)
P/S: mk mới lớp 8 nên cx ko bít lm đúng hay sai, bn tham khảo thôi nhé
\(\frac{xy}{z}+\frac{yz}{x}\ge2y\) ; \(\frac{xy}{z}+\frac{zx}{y}\ge2x\); \(\frac{yz}{x}+\frac{zx}{y}\ge2z\)
Cộng vế với vế:
\(2\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\ge2\left(x+y+z\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
Áp dụng bdt Cosi ta đc:
\(x\ge2\sqrt{\frac{a}{b}}-1\) làm tt rồi nhân xy+yz+zx là ra;x=y=z=1
Lời giải:
Với điều kiện đã cho ta có:
\(F=xy+yz+xz+2xyz=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ac}{(b+c)(a+b)}+\frac{2abc}{(b+c)(c+a)(a+b)}\)
\(=\frac{ab(a+b)+bc(b+c)+ac(a+c)+2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b+c)+bc(b+c+a)+ac(a+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+b+c)(ab+bc)+ac(a+c)}{(a+b)(b+c)(c+a)}=\frac{(a+c)(ba+b^2+bc+ac)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)[b(a+b)+c(b+a)]}{(a+b)(b+c)(c+a)}=\frac{(a+c)(b+c)(a+b)}{(a+b)(b+c)(c+a)}=1\)
Xét: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{bc\left(b-c\right)+ca\left(c-a\right)+ab\left(a-b\right)}{abc}\)
\(=\frac{b^2c-bc^2+ca\left(c-a\right)+a^2b-ab^2}{abc}=\frac{b^2\left(c-a\right)+ca\left(c-a\right)-b\left(c^2-a^2\right)}{abc}\)
\(=\frac{\left(c-a\right)\left(b^2+ca\right)-b\left(c-a\right)\left(c+a\right)}{abc}=\frac{\left(c-a\right)\left(b^2+ca-bc-ba\right)}{abc}\)
\(=\frac{\left(c-a\right)\left(b-a\right)\left(b-c\right)}{abc}=-\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{abc}=-\frac{1}{xyz}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{-1}{xyz}\Leftrightarrow xy+yz+zx=-1\)
\(xy+yz+zx=\frac{a}{b-c}.\frac{b}{c-a}+\frac{b}{c-a}.\frac{c}{a-b}+\frac{c}{a-b}.\frac{a}{b-c}\)\(=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{b\left(a^2-c^2\right)+b^2\left(c-a\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(c-a\right)\left(b^2+ca-ab-bc\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(c-a\right)\left(b\left(b-a\right)+c\left(a-b\right)\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a-b\right)\left(c-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)