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\(P=\dfrac{1}{6-4a}+\dfrac{4}{4a}\ge\dfrac{\left(1+2\right)^2}{6-4a+4a}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(\dfrac{6-4a}{1}=\dfrac{4a}{2}\Rightarrow a=1\)
\(P=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{2a}{b}-\dfrac{2b}{a}-1\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Áp dụng BĐT Minicopski, ta có:
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)
Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)
Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)
\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)
\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
\(M=1+\dfrac{1}{a^2}+\dfrac{2}{a}+1+\dfrac{1}{b^2}+\dfrac{2}{b}=2+2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
Theo BĐT Cauchy-Swarch ta có
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2.\dfrac{4}{a+b}=8\)
áp dụng BĐT AM-GM ta có
\(\dfrac{1}{a^2}+4\ge2\sqrt{\dfrac{1}{a^2}.4}=\dfrac{4}{a}\) ; \(\dfrac{1}{b^2}+4\ge2\sqrt{\dfrac{1}{b^2}.4}=\dfrac{4}{b}\)
Cộng hai vế BĐT trên lại ta được
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+8\ge\dfrac{4}{a}+\dfrac{4}{b}=4\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge16\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge16-8=8\)
\(\Rightarrow M\ge2+8+8=18\) vậy MinM=18 tại x=y=1/2
1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Áp dụng BĐT Cô-si:
\(\dfrac{a^2+1}{a}+\dfrac{a}{a^2+1}\ge2\sqrt{\dfrac{\left(a^2+1\right).a}{a.\left(a^2+1\right)}}=2\)
Vậy Pmin=2\(\Leftrightarrow\dfrac{a^2+1}{a}=\dfrac{a}{a^2+1}\)
\(\Rightarrow a^4+2a^2+1-a^2=0\)
\(\Leftrightarrow\left(a^2+1-a\right)\left(a^2+1+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2-a+1=0\\a^2+a+1=0\end{matrix}\right.\)(vô nghiệm)
Vậy Pmin=2.