cho a>=1,b>=1.CMR a căn 2 b-1+b căn 2 a-1<=ab
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Áp dụng BĐT Bunhiacopxki:
\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(1.a+4.\frac{1}{b}\right)^2\)\(\Rightarrow a^2+\frac{1}{b^2}\ge\frac{1}{17}\left(a+\frac{4}{b}\right)^2\)
\(\Rightarrow\sqrt{a^2+\frac{1}{b^2}}\ge\frac{1}{\sqrt{17}}\left(a+\frac{4}{b}\right)\)
Tương tự, ta có: \(\sqrt{b^2+\frac{1}{c^2}}\ge\frac{1}{\sqrt{17}}\left(b+\frac{4}{c}\right)\)
và \(\sqrt{c^2+\frac{1}{a^2}}\ge\frac{1}{\sqrt{17}}\left(c+\frac{4}{a}\right)\)
Cộng từng vế của các BĐT trên, ta được:
\(P\ge\frac{1}{\sqrt{17}}\left(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)\)\(\ge\frac{1}{\sqrt{17}}\left(a+b+c+\frac{36}{a+b+c}\right)\)(svac - xơ)
\(=\frac{1}{\sqrt{17}}\left[\left(a+b+c\right)+\frac{9}{4\left(a+b+c\right)}+\frac{135}{4\left(a+b+c\right)}\right]\ge\frac{3\sqrt{17}}{2}\)
Vậy \(P=\sqrt{a^2+\frac{1}{b^2}}\)\(+\sqrt{b^2+\frac{1}{c^2}}\)\(+\sqrt{c^2+\frac{1}{a^2}}\ge\frac{3\sqrt{17}}{2}\)
(Dấu "="\(\Leftrightarrow a=b=c=2\))
Bài em làm ok rồi nhưng mà dấu bằng xảy ra bị sai. Em kiểm tra lại!๖²⁴ʱČøøℓ ɮøү 2к⁷༉
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Xét \(\sqrt{a^2-ab+b^2}\) = \(\sqrt{\left(a^2+2ab+b^2\right)-3ab}\) = \(\sqrt{\left(a+b\right)^2-3ab}\)
>= \(\sqrt{\left(a+b\right)^2-\frac{3}{4}\left(a+b\right)^2}\)( bđt ab <= (a+b)^2/4) = 1/2 (a+b)
Tương tự căn (b^2-bc+c^2) >= 1/2(b+c) ; (c^2-ca+a^2) >= 1/2 (c+a)
=> B >= 1/2 . (a+b+b+c+c+a) = 1/2 . 2 . (a+b+c) = 1 => ĐPCM
Dấu "=" xảy ra <=> a=b=c=1/3
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1.
\(\left(1+a\right)^2=\left(1.1+\sqrt{\frac{a}{b}}.\sqrt{ab}\right)^2\le\left(1+\frac{a}{b}\right)\left(1+ab\right)=\frac{\left(a+b\right)\left(1+ab\right)}{b}\)
\(\Rightarrow\frac{1}{\left(1+a\right)^2}\ge\frac{b}{\left(a+b\right)\left(1+ab\right)}\)
\(\left(1+b\right)^2\le\frac{\left(a+b\right)\left(1+ab\right)}{a}\Rightarrow\frac{1}{\left(1+b\right)^2}\ge\frac{a}{\left(a+b\right)\left(1+ab\right)}\)
\(\Rightarrow\frac{1}{\left(1+a\right)^2}+\frac{1}{\left(1+b\right)^2}\ge\frac{a}{\left(a+b\right)\left(1+ab\right)}+\frac{b}{\left(a+b\right)\left(1+ab\right)}=\frac{1}{1+ab}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=1\)
2.
\(P=\sqrt{\frac{a^2}{a^4+3}}+\sqrt{\frac{b^2}{b^4+3}}\le\sqrt{2\left(\frac{a^2}{a^4+3}+\frac{b^2}{b^4+3}\right)}\)
Đặt \(\left(a^2;b^2\right)=\left(x;y\right)\Rightarrow xy=1\)
\(Q=\frac{x}{x^2+3}+\frac{y}{y^2+3}=\frac{x}{x^2+3}+\frac{x}{3x^2+1}-\frac{1}{2}+\frac{1}{2}\)
\(Q=\frac{-\left(x-1\right)^2\left(3x^2-2x+3\right)}{2\left(x^2+3\right)\left(3x^2+1\right)}+\frac{1}{2}\le\frac{1}{2}\)
\(\Rightarrow P\le\sqrt{2Q}\le1\)
\(P_{max}=1\) khi \(a=b=1\)
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Bài 6:
Để B là số nguyên thì \(\sqrt{x}-2+3⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;3\right\}\)
hay \(x\in\left\{9;1;25\right\}\)
trần đắc lợi lần sau nhớ gõ latex nha bạn, như này người làm dễ bị sai đề lắm
\(a\sqrt{b-1}+b\sqrt{a-1}\)
Áp dụng AM-GM :
\(a\sqrt{b-1}+b\sqrt{a-1}\)
\(=a\sqrt{1\cdot\left(b-1\right)}+b\sqrt{1\cdot\left(a-1\right)}\le a\cdot\frac{1+b-1}{2}+b\cdot\frac{1+a-1}{2}\)
\(=\frac{ab}{2}+\frac{ab}{2}=ab\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)
cảm ơn nha