\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)

Ta có:

\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz và AM-GM:

$M=\frac{b^2+c^2}{a^2}+a^2(\frac{1}{b^2}+\frac{1}{c^2})$

$\geq \frac{b^2+c^2}{a^2}+a^2.\frac{4}{b^2+c^2}$

$=(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2})+\frac{3a^2}{b^2+c^2}$

$\geq \sqrt{\frac{b^2+c^2}{a^2}.\frac{a^2}{b^2+c^2}}+\frac{3(b^2+c^2)}{b^2+c^2}$

$=2+3=5$

Vậy $M_{\min}=5$ 

Lời giải:

Áp dụng BĐT AM-GM:

\(\frac{a^4}{(a+2)(b+2)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\geq 4\sqrt[4]{\frac{a^4}{27.27.9}}=\frac{4a}{9}\)

\(\frac{b^4}{(b+2)(c+2)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\geq \frac{4b}{9}\)

\(\frac{c^4}{(c+2)(a+2)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\geq \frac{4c}{9}\)

Cộng theo vế và rút gọn:

\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}+\frac{2(a+b+c)}{27}+\frac{7}{9}\geq\frac{4(a+b+c)}{9}\)

\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}\geq \frac{10(a+b+c)}{27}-\frac{7}{9}=\frac{30}{27}-\frac{7}{9}=\frac{1}{3}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

AM-GM là gì z bn

Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).

Áp dụng bđt Schwars ta có:

\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).

Đẳng thức xảy ra khi a = b = c = 1.

Đẳng thức quen thuộc: \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\) và tương tự cho các mẫu số còn lại

Ta có:

\(\sum\dfrac{1}{a^2+1}=\sum\dfrac{1}{\left(a+b\right)\left(a+c\right)}=\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Mặt khác:

\(2\left(ab+bc+ca\right)\left(a+b+c\right)=\left[a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right]\left(a+b+c\right)\)

\(\ge\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\) (Bunhiacopxki)

\(\Rightarrow\sum\dfrac{1}{a^2+1}\ge\dfrac{\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\left(\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)^2\)

\(=\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^2\)

Do đó ta chỉ cần chứng minh:

\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{3}{2}\)

Đúng theo AM-GM:

\(\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)