Áp dụng BĐT Bunhiacopski:

\(\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\le\left(1^2+1^2\right)\left(a+1+b+1\right)=2\left(a+b+2\right)\\ \Leftrightarrow a+b+2\ge\dfrac{16}{2}=8\\ \Leftrightarrow a+b\ge6\)

Áp dụng BĐT: \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)

\(\Leftrightarrow P=a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left[\dfrac{\left(a+b\right)^2}{2}\right]^2}{2}=\dfrac{\left(a+b\right)^4}{8}\ge\dfrac{6^4}{8}=162\)

Do đó \(P_{min}=162\Leftrightarrow a=b=3\)

tích mình với

ai tích mình

mình tích lại

thanks

Tích mình đi mình tích lại

ra r nha

Như nào vậy?

\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)

Vì \(a=\left(a-1\right)+1\ge2\sqrt{\left(a-1\right).1}=2\sqrt{a-1}\)

\(b=\left(b-4\right)+4\ge2\sqrt{\left(b-4\right).4}=4\sqrt{b-4}\)

\(c=\left(c-9\right)+9\ge2\sqrt{\left(c-9\right).9}=6\sqrt{c-9}\)

=>\(P\le\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)

P max = 11/12 khi a=2; b=8; c =18

Đặt a - 1 = x > 0; b - 1 = y > 0

\(A=\frac{\left(x+1\right)^2}{x}+\frac{\left(y+1\right)^2}{y}\\ A=\frac{x^2+2x+1}{x}+\frac{y^2+2y+1}{y}\\ A=\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+4\)

Với x > 0; y > 0, theo BĐT AM-GM ta có:

\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}\Rightarrow x+\frac{1}{x}\ge2\)

\(y+\frac{1}{y}\ge2\sqrt{y.\frac{1}{y}}\Rightarrow y+\frac{1}{y}\ge2\)

\(\Rightarrow A\ge8\)

Dấu "=" xảy ra khi và chỉ khi x = y = 1 => a = b = 2

Vậy...

Ta có: \(4=\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\)

\(\le\frac{a+b}{2}+\frac{a+1}{2}+\frac{b+1}{2}+1\Rightarrow a+b\ge2\)

Do đó \(P=\frac{a^2}{b}+\frac{b^2}{a}\ge\frac{\left(a+b\right)^2}{a+b}=a+b\ge2\)

Dấu bằng xảy ra khi a = b = 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 bđt bunhiacopski có:

\(\left(a^4+1\right)\left(1+4^2\right)\ge\left(a^2+4\right)^2\)

=> \(\sqrt{a^4+1}\ge\sqrt{\frac{\left(a^2+4\right)^2}{1+4^2}}=\frac{a^2+4}{\sqrt{17}}\)(1)

Tương tự cx có: \(\sqrt{b^4+1}\ge\frac{b^2+4}{\sqrt{17}}\) (2)

Từ (1),(2) => \(F\ge\frac{a^2+b^2+8}{\sqrt{17}}\)

Có (a+2)(b+2)=\(\frac{25}{4}\)

=> \(ab+2a+2b+4=\frac{25}{4}\) <=> \(ab+2a+2b=\frac{9}{4}\)

Áp dụng cosi có:

\(ab\le\frac{a^2+b^2}{2}\)

\(2a\le2\left(a^2+\frac{1}{4}\right)\)

\(2b\le2\left(b^2+\frac{1}{4}\right)\)

=> \(\frac{a^2+b^2}{2}+2a^2+\frac{1}{2}+2b^2+\frac{1}{2}\ge ab+2a+2b=\frac{9}{4}\)

<=> \(\frac{a^2+b^2+4a^2+4b^2}{2}\ge\frac{9}{4}-\frac{1}{2}-\frac{1}{2}=\frac{5}{4}\)

<=> \(\frac{5\left(a^2+b^2\right)}{2}\ge\frac{5}{4}\)

<=> \(a^2+b^2\ge\frac{1}{2}\)

Thay \(a^2+b^2\ge\frac{1}{2}\) vào F có:

\(F\ge\frac{\frac{1}{2}+8}{\sqrt{17}}\)

<=> F \(\ge\frac{\sqrt{17}}{2}\)

Dấu "=" xảy ra <=>\(a=b=\frac{1}{2}\)