\(\dfrac{a}{\sqrt{b^3+1}}=\dfrac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\ge\dfrac{2a}{b+1+b^2-b+1}=\dfrac{2a}{b^2+2}\)

Tương tự và cộng lại:

\(VT\ge\dfrac{2a}{b^2+2}+\dfrac{2b}{c^2+2}+\dfrac{2c}{a^2+2}=a-\dfrac{ab^2}{b^2+2}+b-\dfrac{bc^2}{c^2+2}+c-\dfrac{ca^2}{a^2+2}\)

\(VT\ge6-\left(\dfrac{ab^2}{b^2+2}+\dfrac{bc^2}{c^2+2}+\dfrac{ca^2}{c^2+2}\right)\)

Ta có:

\(\dfrac{ab^2}{b^2+2}=\dfrac{2ab^2}{2b^2+4}=\dfrac{2ab^2}{b^2+b^2+4}\le\dfrac{2ab^2}{3\sqrt[3]{4b^4}}=\dfrac{a}{3}\sqrt[3]{2b^2}=\dfrac{a}{3}\sqrt[3]{2.b.b}\le\dfrac{a}{9}\left(2+b+b\right)\)

Tương tự và cộng lại:

\(VT\ge6-\left(\dfrac{2a}{9}\left(b+1\right)+\dfrac{2b}{9}\left(c+1\right)+\dfrac{2c}{9}\left(a+1\right)\right)\)

\(=6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{9}\left(ab+bc+ca\right)\ge6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{27}\left(a+b+c\right)^2=2\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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

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

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

\(ab+bc+ca\le1\)

\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)

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

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)

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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Bài 2:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:

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

Thiết lập các BĐT tương tự:

\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)

Dấu "=" không xảy ra nên ta có ĐPCM

Lưu ý: lần sau đăng từng bài 1 thôi nhé !

1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)

TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)

\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)

Cộng vế với vế ta được:

\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)

Đặt vế trái là T, ta có:

\(\dfrac{a}{\sqrt{b+1}}=\dfrac{a\sqrt{2}}{\sqrt{2}.\sqrt{b+1}}\ge\dfrac{a\sqrt{2}}{\dfrac{b+1+2}{2}}=\dfrac{a.2\sqrt{2}}{b+3}\)

Tương tự: \(\dfrac{b}{\sqrt{c+1}}\ge\dfrac{b.2\sqrt{2}}{c+3}\)

\(\dfrac{c}{\sqrt{a+1}}\ge\dfrac{c.2\sqrt{2}}{a+3}\)

Cộng vế theo vế các BĐT vừa chứng minh, ta được

\(T\ge2\sqrt{2}\left(\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{a+3}\right)=2\sqrt{2}\left(\dfrac{a^2}{ab+3a}+\dfrac{b^2}{bc+3b}+\dfrac{c^2}{ac+3c}\right)\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{3^2}{\dfrac{3^2}{3}+9}=\dfrac{3\sqrt{2}}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1

b) Đặt vế trái là N,ta có:

\(\sum\sqrt{\dfrac{a^3}{b+3}}=\sum\sqrt{\dfrac{a^4}{ab+3}}=\sum\dfrac{a^2}{\sqrt{ab+3}}=\sum\dfrac{2a^2}{\sqrt{4a\left(b+3\right)}}\ge\sum\dfrac{2a^2}{\dfrac{4a+b+3}{2}}=\sum\dfrac{4a^2}{4a+b+3}\)

\(\sum\dfrac{4a^2}{4a+b+3}\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+b+3+4b+c+3+4c+a+3}=\dfrac{3}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1