Đề bài sai

Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

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

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)

Ta có : \(\sqrt{\dfrac{a^2}{b}}+\sqrt{\dfrac{b^2}{a}}\)

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

\(=\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)

Theo BĐT Cô Si dưới dạng engel ta có :

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

Dấu \("="\) xảy ra khi \(a=b\)

Chúc bạn học tốt

Đề thiếu a,b > 0

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

\(\Leftrightarrow\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)

Áp dụng bđt Svacxo, ta có:

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

=> ĐPCM

Áp dụng AM-GM: \(a\sqrt{b-1}+b\sqrt{a-1}\le a.\dfrac{b-1+1}{2}+b.\dfrac{a-1+1}{2}=ab\)

\(VT\ge\dfrac{6}{ab}+\sqrt{3ab+4}\)

( dự đoán dấu = xảy ra khi a=b=2)

Áp dụng cauchy-schwarz:

\(\dfrac{6}{ab}=\dfrac{18}{3ab}+\dfrac{2}{4}-\dfrac{1}{2}\ge\dfrac{\left(\sqrt{18}+\sqrt{2}\right)^2}{3ab+4}-\dfrac{1}{2}=\dfrac{32}{3ab+4}-\dfrac{1}{2}\)

Áp dụng AM-GM một lần nữa:

\(VT\ge\dfrac{32}{3ab+4}+\sqrt{3ab+4}-\dfrac{1}{2}=\dfrac{32}{3ab+4}+\dfrac{\sqrt{3ab+4}}{2}+\dfrac{\sqrt{3ab+4}}{2}-\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{32}{4}}-\dfrac{1}{2}=\dfrac{11}{2}\)

Dấu = xảy ra khi a=b=2

P/s: Nothing

a: \(a+\dfrac{1}{a}\ge2\sqrt{a\cdot\dfrac{1}{a}}=2\)

b: \(\Leftrightarrow\dfrac{a^2+a+1+1}{\sqrt{a^2+a+1}}>=2\)

=>\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\)(1)

\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\sqrt{\sqrt{a^2+a+1}\cdot\dfrac{1}{\sqrt{a^2+a+1}}}=2\)

nên (1) đúng

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.