Áp dụng BĐT đã chứng minh ở phần trước:

\(\left(a+b+c\right)^2\le2k\left(1+bc\right)^2=4\left(1+bc\right)^2\)

\(\Leftrightarrow a^2\left(a+b+c\right)^2\le4a^2\left(1+bc\right)^2\)

\(\Rightarrow a\left(a+b+c\right)\le2a\left(1+bc\right)\)

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

Hoàn toàn tương tự, ta có: \(\frac{b}{1+ac}\le\frac{2b}{a+b+c}\) ; \(\frac{c}{1+ca}\le\frac{2c}{a+b+c}\)

Cộng vế với vế: \(P\le2\)

\(P_{max}=2\) khi \(\left(a;b;c\right)=\left(0;1;1\right)\) và hoán vị

\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)

\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)

=>P<=1/a+1/b+1/c=3

Dấu = xảy ra khi a=b=c=1

Cái c là \(\dfrac{2}{\sqrt{1+c^2}}\) ạ

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

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

Từ điều kiện \(ab+bc+ca=1\), đặt \(\left\{{}\begin{matrix}a=tanx\\b=tany\\c=tanz\end{matrix}\right.\) với \(x+y+z=\dfrac{\pi}{2}\)

Xét \(Q=\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}=\dfrac{1}{1+tan^2x}+\dfrac{1}{1+tan^2y}+\dfrac{1}{\sqrt{1+tan^2z}}\)

\(Q=cos^2x+cos^2y+cosz=1+\dfrac{1}{2}\left(cos2x+cos2y\right)+cosz\)

\(=1+cos\left(x+y\right)cos\left(x-y\right)+cosz\le1+cos\left(x+y\right)+cosz\)

\(=1+cos\left(\dfrac{\pi}{2}-z\right)+cosz=1+sinz+cosz=1+\sqrt{2}sin\left(z+\dfrac{\pi}{4}\right)\le1+\sqrt{2}\)

\(\Rightarrow P\le2\left(1+\sqrt{2}\right)-2=2\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=y=\dfrac{\pi}{8}\\z=\dfrac{\pi}{4}\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\sqrt{2}-1;\sqrt{2}-1;1\right)\)

a/ \(k+2bc=a^2+\left(b^2+c^2+2bc\right)=a^2+\left(b+c\right)^2\ge2a\left(b+c\right)\)

Đẳng thức xảy ra khi \(b+c=a\)

b/ BĐT cần chứng minh tương đương:

\(k+2\left(ab+bc+ca\right)\le2k\left(1+2bc+2b^2c^2\right)\)

\(\Leftrightarrow4kb^2c^2+4kbc+k\ge2a\left(b+c\right)+2bc\)

Điều này hiển nhiên đúng do: \(k\ge1\Rightarrow4kbc\ge4bc\)

\(\Rightarrow4kb^2c^2+4kbc+k\ge4bc+k=2bc+\left(k+2bc\right)\ge2bc+2a\left(b+c\right)\)

c/ BĐT đã cho sai.

Phản ví dụ: giả sử cho \(a^2+b^2+c^2=2\)

BĐT trở thành \(a+b+c+abc\le2\)

Nhưng với \(a=b=c=\sqrt{\frac{2}{3}}\) thì \(a+b+c+abc=\frac{11\sqrt{6}}{9}>2\)

Bài tập áp dụng:

Áp dụng BĐT ở câu b:

\(\left(a+b+c\right)^2\le2k\left(1+bc\right)^2\Rightarrow\frac{a^2}{\left(a+b+c\right)^2}\ge\frac{a^2}{2k\left(1+bc\right)^2}\)

\(\Leftrightarrow\frac{a}{1+bc}\le\sqrt{2k}.\frac{a}{a+b+c}=\sqrt{2}.\frac{a}{a+b+c}\)

Hoàn toàn tương tự, ta có:

\(\frac{b}{1+ca}\le\sqrt{2}.\frac{b}{a+b+c}\) ; \(\frac{c}{1+ca}\le\sqrt{2}.\frac{c}{a+b+c}\)

Cộng vế với vế: \(P\le\sqrt{2}.\frac{a+b+c}{a+b+c}=\sqrt{2}\)

\(P_{max}=\sqrt{2}\)khi \(\left(a;b;c\right)=\left(0;\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}\right)\) và hoán vị

Lại có:

\(a\left(1+bc\right)\le a\left(1+\frac{b^2+c^2}{2}\right)=a\left(\frac{3-a^2}{2}\right)-1+1=-\frac{1}{2}\left(a-1\right)^2\left(a+2\right)+1\le1\)

\(\Rightarrow a^2\left(1+bc\right)\le a\Rightarrow\frac{a}{1+bc}\ge a^2\)

Tương tự ta có: \(\frac{b}{1+ca}\ge b^2\) ; \(\frac{c}{1+ab}\ge c^2\)

Cộng vế với vế: \(P\ge1\Rightarrow P_{min}=1\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị

c/

Nếu dấu là trừ:

BĐT cần chứng minh tương đương:

\(\left(a+b+c-\frac{2}{k}abc\right)^2\le2k\)

Ta có:

\(VT=\left[\left(a+b\right).1+c\left(1-\frac{2}{k}ab\right)\right]^2\)

\(VT\le\left[\left(a+b\right)^2+c^2\right]\left[1+\left(1-\frac{2}{k}ab\right)^2\right]\)

\(VT\le\left(k+2ab\right)\left(2-\frac{4}{k}ab+\frac{4a^2b^2}{k^2}\right)\)

\(VT\le2k-\frac{4}{k}a^2b^2+\frac{8}{k^2}\left(ab\right)^3\)

Do đó ta chỉ cần chứng minh: \(2k-\frac{4}{k}\left(ab\right)^2+\frac{8}{k^2}\left(ab\right)^3\le2k\)

\(\Leftrightarrow\frac{1}{k}\left(ab\right)^2-\frac{2}{k^2}\left(ab\right)^3\ge0\)

\(\Leftrightarrow\frac{1}{k}\left(ab\right)^2\left(1-\frac{2ab}{k}\right)\ge0\)

Từ giả thiết \(k=a^2+b^2+c^2\ge a^2+b^2\ge2ab\Rightarrow\frac{2ab}{k}\le1\)

\(\Rightarrow1-\frac{2ab}{k}\ge0\Rightarrow\frac{1}{k}\left(ab\right)^2\left(1-\frac{2ab}{k}\right)\ge0\) (đpcm)

\(c\left(1+ab\right)\le c\left(1+\dfrac{a^2+b^2}{2}\right)=c\left(1+\dfrac{1-c^2}{2}\right)=1-\dfrac{1}{2}\left(c-1\right)^2\left(c+2\right)\le1\)

\(\Rightarrow c^2\left(1+ab\right)\le c\Rightarrow\dfrac{c}{1+ab}\ge c^2\)

Hoàn toàn tương tự ta có: \(\dfrac{a}{1+bc}\ge a^2\) ; \(\dfrac{b}{1+ac}\ge b^2\)

Cộng vế: \(VT\ge a^2+b^2+c^2=1\) (đpcm)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị

Cách 2:

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}[a(1+bc)+b(1+ac)+c(1+ab)]\geq (a+b+c)^2\)

\(\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{a+b+c+3abc}\)

 Ta sẽ CM: 

\(\frac{(a+b+c)^2}{a+b+c+3abc}\geq 1\)

\(\Leftrightarrow 1+2(ab+bc+ac)\geq a+b+c+3abc\)

Vì $a^2+b^2+c^2=1\Rightarrow a,b,c\leq 1$

$\Rightarrow (a-1)(b-1)(c-1)\leq 0$

$\Leftrightarrow 1+ ab+bc+ac\geq a+b+c+abc(1)$

Áp dụng BĐT AM-GM:

$ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}\geq 3\sqrt[3]{a^2b^2c^2.abc}=3abc\geq 2abc(2)$

Từ $(1);(2)\Rightarrow 1+2(ab+bc+ac)\geq a+b+c+3abc$

Ta có đpcm

Dấu "=" xảy ra khi $(a,b,c)=(1,0,0)$ và hoán vị.

\(a\text{) }\)Áp dụng: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (a, b > 0). Dấu "=" xảy ra khi a = b.

\(\frac{1}{a^2+b^2}+\frac{1}{ab}=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2.\frac{\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\)

\(=6\left[\frac{1}{\left(a+b\right)^2}+\frac{27}{8}\left(a+b\right)+\frac{27}{8}\left(a+b\right)\right]-\frac{81}{2}\left(a+b\right)\)

\(\ge6.3\sqrt[3]{\frac{1}{\left(a+b\right)^2}.\frac{27}{8}\left(a+b\right).\frac{27}{8}\left(a+b\right)}-\frac{81}{2}\left(a+b\right)\)

\(=\frac{81}{2}-\frac{81}{2}\left(a+b\right)\)

Tương tự: \(\frac{1}{b^2+c^2}+\frac{1}{bc}\ge\frac{81}{2}-\frac{81}{2}\left(b+c\right)\)

\(\frac{1}{c^2+a^2}+\frac{1}{ca}\ge\frac{81}{2}-\frac{81}{2}\left(c+a\right)\)

Cộng theo vế ta được 

\(A\ge3.\frac{81}{2}-81\left(a+b+c\right)=3.\frac{81}{2}-81=\frac{81}{2}\)

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

Vậy GTNN của A là \(\frac{81}{2}.\)

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)