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+)\(\frac{3}{4}\ge a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow\frac{1}{8}\ge abc\)
+) \(P=8abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(32abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-24abc\)
\(\ge4\sqrt[4]{\frac{32}{abc}}-24abc\ge4\sqrt[4]{\frac{32}{\frac{1}{8}}}-3=16-3=13\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{2}\)
\(A=\frac{19}{ab}+\frac{6}{a^2+b^2}+2018\left(a^4+b^4\right)\)
\(=6\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{16}{ab}+2018\left(a^4+b^4\right)\)
\(\ge\frac{24}{\left(a+b\right)^2}+\frac{64}{\left(a+b\right)^2}+\frac{2018\left(a+b\right)^4}{8}=24+64+\frac{2018}{8}=\frac{1361}{4}\)
Vậy GTNN của A là \(\frac{1361}{4}\) khi \(a=b=\frac{1}{2}\)
a)\(B=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)
Áp dụng BĐT AM-GM ta có:
\(B=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)
\(\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{1}{2ab}\cdot8ab}-\left(a+b\right)^2=7\)
Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)
Vậy \(Min_B=7\) khi \(a=b=\frac{1}{2}\)
b)\(C\ge\frac{1}{1-3ab\left(a+b\right)}+\frac{4}{ab\left(a+b\right)}\)
\(\ge\frac{16}{1-3ab\left(a+b\right)+3ab\left(a+b\right)}+\frac{1}{\frac{\left(a+b\right)^3}{4}}\ge16+4=20\)
Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)
Vậy \(Min_C=20\) khi \(a=b=\frac{1}{2}\)
1. Cho a,b,c thực dương thỏa mãn: abc=1
\(b^4+c^4-bc\left(b^2+c^2\right)=\left(b^2+bc+c^2\right)\left(b-c\right)^2\)
\(\Rightarrow b^4+c^4\ge bc\left(b^2+c^2\right)\)
Tương tự\(\Rightarrow\Sigma_{cyc}\frac{a}{a+b^4+c^4}\le\Sigma_{cyc}\frac{a}{a+bc\left(b^2+c^2\right)}=\Sigma_{cyc}\frac{a}{bc\left(a^2+b^2+c^2\right)}=\frac{1}{a^2+b^2+c^2}\Sigma_{cyc}\frac{a}{bc}\)
\(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}=a^2+b^2+c^2\)
\(\Rightarrow\frac{1}{a^2+b^2+c^2}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)=1\)
oke rồi he
@Nub :v
Áp dụng Bunhiacopski ta dễ có:
\(\frac{a}{b^4+c^4+a}=\frac{a\left(1+1+a^3\right)}{\left(b^4+c^4+a\right)\left(1+1+a^3\right)}\le\frac{a^4+2a}{\left(a^2+b^2+c^2\right)^2}\)
Tương tự:
\(\frac{b}{a^4+c^4+b}\le\frac{b^4+2b}{\left(a^2+b^2+c^2\right)^2};\frac{c}{a^4+b^4+c}\le\frac{c^4+2c}{\left(a^2+b^2+c^2\right)^2}\)
Cộng lại:
\(A\le\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\)
Ta đi chứng minh:
\(\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\le1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
Cái này luôn đúng theo Cauchy
Đẳng thức xảy ra tại a=b=c=1
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow17\cdot\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow\sqrt{17}\cdot\sqrt{a^2+\frac{1}{b^2}}\ge a+\frac{4}{b}\)
Tương tự ta có :
\(\sqrt{17}\cdot\sqrt{b^2+\frac{1}{c^2}}\ge b+\frac{4}{c}\)
\(\sqrt{17}\cdot\sqrt{c^2+\frac{1}{a^2}}\ge c+\frac{4}{a}\)
Cộng theo vế của 3 bđt ta được :
\(\sqrt{17}\cdot\left(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\right)\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(\Leftrightarrow\sqrt{17}\cdot A\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
Áp dụng bất đẳng thức Cô-si :
\(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(=16a+\frac{4}{a}+16b+\frac{4}{b}+16c+\frac{4}{c}-15a-15b-15c\)
\(\ge2\sqrt{\frac{4\cdot16a}{a}}+2\sqrt{\frac{4\cdot16b}{b}}+2\sqrt{\frac{4\cdot16c}{c}}-15\left(a+b+c\right)\)
\(\ge16+16+16-15\cdot\frac{3}{2}=\frac{51}{2}\)
Do đó : \(\sqrt{17}\cdot A\ge\frac{51}{2}\)
\(\Leftrightarrow A\ge\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)
\(P=\frac{16a}{3}+\frac{1}{b}+\frac{4}{4c}\ge\frac{16a}{9}+\frac{16a}{9}+\frac{16a}{9}+\frac{9}{b+4c}\ge4\sqrt[4]{\frac{4096}{81}.\frac{a^3}{b+4c}}=\frac{32}{3}\)
"=" \(\Leftrightarrow\)\(\left(a;b;c\right)=\left(\frac{3}{2};\frac{9}{8};\frac{9}{16}\right)\)
\(P=\left(\frac{a}{b}+\frac{b}{a}\right)^2-\left(\frac{a}{b}+\frac{b}{a}\right)-1\)
Đặt \(t=\frac{a}{b}+\frac{b}{a}\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\)
\(f\left(t\right)=t^2-t-1\)
\(-\frac{b}{2a}=\frac{1}{2}\notin(-\infty;-2]\cup[2;+\infty)\)
\(f\left(-2\right)=5\) ; \(f\left(2\right)=1\)
\(\Rightarrow P_{min}=1\) khi \(t=2\Leftrightarrow\frac{a}{b}+\frac{b}{a}=2\Leftrightarrow a=b\)
sao lại -1 vậy bn ?