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1: ĐKXĐ: a,b>0, a\(\ne b\)
\(\Rightarrow Q=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^3+2a\sqrt{a}+b\sqrt{b}}{3\sqrt{a}\left(a\sqrt{a}+b\sqrt{b}\right)}+\dfrac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}=\dfrac{a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}-b\sqrt{b}+2a\sqrt{a}+b\sqrt{b}}{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) \(=\dfrac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=0\)
\(\Rightarrow Q\) ko phụ thuộc vào a,b Vậy...
2: Ta có \(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\)
\(\Rightarrow P=\dfrac{x+y}{xy}\cdot\sqrt{x^2y^2+\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}}\ge\dfrac{2\sqrt{xy}}{xy}\cdot\sqrt{17}\cdot\sqrt[34]{\dfrac{x^2y^2}{16^{16}}}=\sqrt{17}\cdot\dfrac{2}{\sqrt{xy}}\cdot\sqrt[17]{\dfrac{xy}{16^8}}\) \(=\sqrt{17}\cdot\sqrt[17]{\dfrac{2^{17}}{\sqrt{x^{17}y^{17}}}\cdot\dfrac{\sqrt{x^2y^2}}{2^{32}}=\sqrt{17}\cdot\sqrt[17]{\dfrac{1}{\sqrt{x^{15}y^{15}}\cdot2^{15}}}\ge\sqrt{17}\cdot\sqrt[17]{\dfrac{1}{\sqrt{\dfrac{1}{4^{15}}}\cdot2^{15}}}=\sqrt{ }17}\)
Dấu = xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\) Vậy...
Bài 286: Bất đẳng thức neibizt khá nổi tiếng :D
Bđt <=> \(\dfrac{a}{b+c}+\dfrac{1}{2}+\dfrac{b}{c+a}+\dfrac{1}{2}+\dfrac{c}{a+b}+\dfrac{1}{2}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(2a+2b+2c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{c+a}+\dfrac{1}{b+c}\right)\ge9\) ( Có thể đơn giản hóa bất đẳng thức bằng việc đặt biến phụ )
Đặt: \(\left\{{}\begin{matrix}x=b+c\\y=c+a\\z=a+b\end{matrix}\right.\) khi đó ta có: \(\left\{{}\begin{matrix}a=\dfrac{y+z-x}{2}\\b=\dfrac{z+x-y}{2}\\c=\dfrac{x+y-z}{2}\end{matrix}\right.\) Bất đẳng thức trở thành: \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\) ( luôn đúng theo AM-GM )
Vậy bất đẳng thức đã được chứng minh. Dấu "=" xảy ra tại a=b=c
C286.(Cách khác)
Áp dụng BĐT BSC và BĐT \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\):
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{a^2}{ab+ca}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{\dfrac{2}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c\)
Ta cần chứng minh \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\ge1\)
\(\Leftrightarrow\dfrac{a^3}{ab+b^2}+\dfrac{b^3}{ab+a^2}\ge1\) \(\Leftrightarrow\dfrac{a^3}{b\cdot\left(a+b\right)}+\dfrac{b^3}{a\left(a+b\right)}\ge1\) \(\Leftrightarrow\dfrac{a^4+b^4}{ab\left(a+b\right)}\ge1\Leftrightarrow\dfrac{a^4+b^4}{a+b}\ge1\)
Áp dụng bđt Cô-si vào 2 số a,b>0 :
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\a^4+b^4\ge2a^2b^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\cdot\left(a^2+b^2\right)\ge\left(a+b\right)^2\\2\cdot\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\end{matrix}\right.\) \(\Rightarrow a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(a+b\right)^4}{8}\)
\(\Rightarrow\dfrac{a^4+b^4}{a+b}\ge\dfrac{\left(a+b\right)^3}{8}\ge\dfrac{\left(2\sqrt{ab}\right)^3}{8}=1\)
Dấu bằng xảy ra \(\Leftrightarrow a=b=1\) Vậy...
Ta có:ab=1⇔a=\(\dfrac{1}{b}\)
Thay a=\(\dfrac{1}{b}\) vào \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\) có
\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^3}{1+\left(\dfrac{1}{b}\right)^2}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^3}{\dfrac{b^2+1}{b^2}}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^5}{1+b^2}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3+b^5}{1+b^2}\)=\(\dfrac{\dfrac{1+b^8}{b^3}}{1+b^2}\)
Mà b là số thực dương nên \(\dfrac{\dfrac{1+b^8}{b^3}}{1+b^2}\)≥1
vậy \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\)≥1
Bài 5
Giả sử diện tích tam giác ABC là số nguyên.
\(\Rightarrow S_{ABC}=\dfrac{1}{2}sinA\cdot AB\cdot AC\) là số nguyên
\(\Rightarrow sinA\cdot AB\cdot AC⋮2\) \(\Rightarrow AB\cdot AC⋮2\)( vì \(sinA< 1\) ) vô lí vì AB,AC,BC đều là số nguyên tố \(\Rightarrow\) giả sử sai Vậy ...
139:
Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)\)
GT \(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3\)
\(\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)\) Tương tự ta được:
\(\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)\)
\(\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)\)
\(\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)\) \(\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1\)
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3\)
câu trả lời :
Đặt x=1a,y=1b,z=1c(a,b,c>0)x=1a,y=1b,z=1c(a,b,c>0)
GT ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3
⇒y2xy2+2x2=1b2:(1ab2+2a2)=1b2:(a+2b2a2b2)=a2a+2b2=a−2ab2a+2b2≥a−2ab23b3√ab=a−233√a2b2≥a−29(a+b+ab)
`\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=6-18x`
`<=>\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=6-18x`
`<=>(9x-3)/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6(3x-1)=0`
`<=>(3x-1)(3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6)=0`
Ta thấy `3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6>0`
`=>3x-1=0`
`=>3x=1`
`=>x=1/3`
Vậy `S={1/3}`
`1/(x^2+9x+20)=1/15-1/(x^2+5x+4)(x ne -1,-4,-5)`
`=>1/((x+4)(x+5))=1/15-1/((x+1)(x+4))`
`=>1/(x+4)-1/(x+5)=1/15-1/((x+1)(x+4))`
`=>3/(x+4)-3/(x+5)=3/15-3/((x+1)(x+4))`
`=>3/(x+4)-3/(x+5)=3/15-1/(x+1)+1/(x+4)`
`=>2/(x+4)-3/(x+5)+1/(x+1)=3/15`
`=>30(x+1)(x+5)-45(x+1)(x+4)+15(x+4)(x+5)=3(x+1)(x+4)(x+5)`
`=>30(x^2+6x+5)-45(x^2+5x+4)+15(x^2+9x+20)=3(x^2+5x+4)(x+5)`
`<=>90x+270=3(x^3+8x^2+29x+20)`
`<=>x^3+24x^2-3x-210=0`
`=>x=-23\or\x=2,85\or\x=-3`
`A=(10^50+2)/(10^50-1)`
`=1+3/(10^50-1)`
Tương tự:
`B=1+3/(10^50-3)`
`10^50-1>10^50-3>0`
`=>3/(10^50-1)<3/(10^50-3)`
`=>A<B`
`20.2^x+1=10.4^2+1`
`=>20.2^x=10.4^2`
`=>2^x=4^2/2=2^3`
`=>x=3`
Vậy x=3
2.
\(\left(a+b\right)^2\ge4ab\ge16\Rightarrow a+b\ge4\)
\(\dfrac{a^2+b^2}{a+b}\ge\dfrac{\left(a+b\right)^2}{2\left(a+b\right)}=\dfrac{a+b}{2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{a+b}{2}\ge\dfrac{6}{a+b-1}\)
\(\Leftrightarrow\left(a+b\right)\left(a+b-1\right)-12\ge0\)
\(\Leftrightarrow\left(a+b-4\right)\left(a+b+3\right)\ge0\) (luôn đúng với mọi \(a+b\ge4\))
Dấu "=" xảy ra khi \(a=b=2\)
Câu cuối:
Ta chứng minh BĐT phụ sau: với mọi x;y;z dương, ta luôn có: \(\dfrac{x^3+y^3}{x^2+y^2}\ge\dfrac{x+y}{2}\)
Thật vậy, bất đẳng thức tương đương:
\(2\left(x^3+y^3\right)\ge\left(x+y\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (đúng)
Áp dụng:
\(P\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\ge6\)
\(P_{min}=6\) khi \(a=b=c=2\)