Áp dụng BĐT phụ:

\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)

P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)

\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)

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

Xét M=\(\sum\dfrac{a}{3a+2b+c}\)

\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)

\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)

Mà

\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)

\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)

\(\Rightarrow\)\(M\le\dfrac{1}{2}\)

\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)

Dấu \(=\) xảy ra khi và chỉ khi x=y=z=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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)

c=c.1 thay 1 bằng a+b+c xong cô si

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

\(\Leftrightarrow\sqrt{\dfrac{97}{4}}P=\sqrt{4+\dfrac{81}{4}}\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{c^2+\dfrac{1}{c^2}}\)

\(\ge\left(2a+\dfrac{9}{2a}\right)+\left(2b+\dfrac{9}{2b}\right)+\left(2c+\dfrac{9}{2c}\right)\)

\(=2\left(a+b+c\right)+\dfrac{9}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\ge4+\dfrac{9}{2}.\dfrac{9}{a+b+c}=4+\dfrac{81}{4}=\dfrac{97}{4}\)

\(\Rightarrow P\ge\sqrt{\dfrac{97}{4}}\)

PS: Lần sau chép đề cẩn thận nhé bạn.

Toán lớp 10 Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH §1. Bất đẳng thức

Nếu là \(\ge \) thì easy rồi. Áp dụng BĐT Min....

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

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

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}\)

\(\ge\sqrt{2^2+\left(\dfrac{9}{2}\right)^2}=\sqrt{\dfrac{97}{4}}=VP\)

Khi \(a=b=c=\frac{2}{3}\)

Do \(a\le\left|a\right|,b\le\left|b\right|\) nên ta chỉ cần chứng minh

\(\dfrac{\left|a\right|}{\sqrt{a^2+b^2}}+\dfrac{\left|b\right|}{\sqrt{9a^2+b^2}}+\dfrac{2\left|a\right|\left|b\right|}{\sqrt{a^2+b^2}.\sqrt{9a^2+b^2}}\le\dfrac{3}{2}\)

Đặt \(a^2=x,b^2=3y^2\)

\(P=2\sqrt{\dfrac{x}{x+3y}}+2\sqrt{\dfrac{y}{y+3x}}+4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le3\)

Sử dụng BĐT AM-GM, ta có

\(2\sqrt{\dfrac{x}{x+3y}}\le\dfrac{x}{x+y}+\dfrac{x+y}{3x+y},2\sqrt{\dfrac{y}{y+3x}}\le\dfrac{y}{x+y}+\dfrac{x+y}{y+3x}\)\(4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}+\dfrac{1}{2}\)

Cộng ba bất đẳng thức trên vế theo vế

\(P\le\dfrac{3}{2}+\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\)

Và do đó chứng minh sẽ hoàn tất nếu ta chỉ ra được rằng:

\(\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\le\dfrac{3}{2}\)

Ta có: \(\dfrac{3}{2}-\dfrac{x+y}{x+3y}-\dfrac{x+y}{y+3x}-\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{3}{2}-\dfrac{4\left(x+y\right)^2+8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{\left(x-y\right)^2}{2\left(x+3y\right)\left(y+3x\right)}\ge0\)Bài toán được chứng minh xong. Đẳng thức xảy ra khi \(b=\sqrt{3}a>0\)