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Ta có: \(\hept{\begin{cases}x^2+y^2=1\\\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\end{cases}}\)
\(\Leftrightarrow b\left(a+b\right)x^4+a\left(a+b\right)y^4=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow b^2x^4+a^2y^4-2abx^2y^2=0\)
\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\)
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}=\frac{y^{2016}}{b^{1008}}=\frac{1}{\left(a+b\right)^{1008}}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{21008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Em vào câu hỏi tương tự tham khảo:
Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)
Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)
<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)
<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)
<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)
<=> \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)
Khi đó: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=2\frac{x^{2016}}{a^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)
1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2\) (do x+y=1)
\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)
Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)
Vậy \(x^3+y^3\ge\dfrac{1}{4}\)
2.
a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))
Đẳng thức xảy ra \(\Leftrightarrow a=b\)
b) Lần trước mk giải rồi nhá
3.
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)
b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)
\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)
\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)
\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow ayz+bxz+cxy=0\left(1\right)\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xyc+ayz+xbz}{abc}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)(đpcm)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ac}\right)\)
\(=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{cxy+ayz+bzx}{abc}\right)\)\(=1-0=1\left(dpcm\right)\)
a) Sai đề.
\(\dfrac{a+b}{b^2}\sqrt[]{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\left|a\right|}{\left|a+b\right|}=\left|a\right|\)
b) Sai đề.
\(\dfrac{a\sqrt[]{b}+b\sqrt[]{a}}{\sqrt[]{ab}}:\dfrac{1}{\sqrt[]{a}-\sqrt[]{b}}=\dfrac{\sqrt[]{ab}\left(\sqrt[]{a}+\sqrt[]{b}\right)}{\sqrt[]{ab}}.\left(\sqrt[]{a}-\sqrt[]{b}\right)=a-b\)