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1/a+a/4+3/4.a
2 cái đầu dùng cosi,,cái 3 dùng a>=2
dấu = khi a=2
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\(A=x-2\sqrt{x}\left(\sqrt{y}+1\right)+\left(\sqrt{y}+1\right)^2+\left(3y+1-\left(\sqrt{y}+1\right)^2\right)\)
\(=\left(\sqrt{x}-\sqrt{y}-1\right)^2+2\left(y-\sqrt{y}+\frac{1}{4}\right)-\frac{1}{2}\)
\(=\left(\sqrt{x}-\sqrt{y}-1\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)
Amin= -1/2 khi y=1/4; x=9/4
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Áp dụng BĐT AM-GM ta có:
\(A=5a+6b+7c+\frac{1}{a}+\frac{8}{b}+\frac{27}{c}\)
\(=4\left(a+b+c\right)+\left(\frac{1}{a}+a\right)+\left(\frac{8}{b}+2b\right)+\left(\frac{27}{c}+3c\right)\)
\(\ge4\cdot6+2\sqrt{\frac{1}{a}\cdot a}+2\sqrt{\frac{8}{b}\cdot2b}+2\sqrt{\frac{27}{c}\cdot3c}\)
\(\ge24+2+2\cdot4+2\cdot9=52\)
Xảy ra khi \(\frac{1}{a}=a;\frac{8}{b}=2b;\frac{27}{c}=3c\Rightarrow a=1;b=2;c=3\)
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Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)
Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$
Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:
$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$
Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$
Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$
$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$
Nên $Q \leq \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)
Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$
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\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)
\(P=\frac{a+b}{2a-b}+\frac{b+c}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{\frac{2ac}{a+c}+c}{2c-\frac{2ac}{a+c}}=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)
Dấu "=" xảy ra khi \(a=b=c\)
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=1+1/a+a/4-a/4>=1+1-1/2=3/2
khi a=2