Đề bài có nhầm lẫn gì ko nhỉ?

\(T=\dfrac{ab}{a^2+b^2+ab}+\dfrac{bc}{b^2+c^2+2bc}+\dfrac{ca}{c^2+a^2+ca}\le\dfrac{ab}{2ab+ab}+\dfrac{bc}{2bc+bc}+\dfrac{ca}{2ca+ca}=1\)

 Bạn tham khảo bài này trên Quanda nha.  

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

\(P\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{2}\left(\dfrac{ab+bc+ca}{abc}\right)\le\dfrac{1}{2}\left(\dfrac{a^2+b^2+c^2}{abc}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Áp dụng cosi:

`a^2+bc>=2a\sqrt{bc}`

Hoàn toàn tương tự:

`=>P<=1/2(1/sqrt{ab}+1/sqrt{bc}+1/sqrt{ca})`

Áp dụng cosi:

`1/a+1/b+1/c>=1/sqrt(ab)+1/sqrt(bc)+1/sqrt(ca)`

`=>P<=1/2(1/a+1/b+1/c)`

`=>P<=1/2((ab+bc+ca)/(abc))<=(a^2+b^2+c^2)/(2(abc))=1/2`

Dấu "=" `<=>a=b=c=3`

Theo tớ là tìm Min chứ nhỉ ??

\(ab\left(a+b\right)=a^2+b^2-ab\Rightarrow ab=\dfrac{a^2+b^2-ab}{a+b}\)

\(A=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)\left(a^2+b^2-ab\right)}{a^3b^3}=\dfrac{\left(a+b\right)ab\left(a+b\right)}{a^3b^3}=\dfrac{\left(a+b\right)^2}{a^2b^2}\)

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

Ta có: \(a^2+b^2-ab>0;\forall a;b\ne0\Rightarrow\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}\ge0\)

\(\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}=\dfrac{a^2+b^2+2ab}{a^2+b^2-ab}=\dfrac{4\left(a^2+b^2-ab\right)-3\left(a^2+b^2-2ab\right)}{a^2+b^2-ab}=4-\dfrac{3\left(a-b\right)^2}{a^2+b^2-ab}\le4\)

\(\Rightarrow0\le\dfrac{\left(a+b\right)^2}{a^2+b^2-ab}\le4\)

\(\Rightarrow A\le16\)

\(A_{max}=16\) khi \(a=b=\dfrac{1}{2}\)

\(ab+bc+ca=3\Rightarrow\left\{{}\begin{matrix}a+b+c\ge3\\abc\le1\end{matrix}\right.\)

Ta sẽ chứng minh \(P\le\dfrac{3}{8}\)

\(P\le\dfrac{a}{6a+2}+\dfrac{b}{6b+2}+\dfrac{c}{6c+2}\) nên chỉ cần chứng minh: \(\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\le\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\ge\dfrac{3}{4}\)

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

\(\Leftrightarrow\dfrac{6\left(a+b+c\right)+30}{27abc+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)

\(\Rightarrow\dfrac{6\left(a+b+c\right)+30}{27+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)

\(\Leftrightarrow24\left(a+b+c\right)+120\ge165+9\left(a+b+c\right)\)

\(\Leftrightarrow a+b+c\ge3\) (đúng)

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

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

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

Đặt \(\left\{{}\begin{matrix}a-2=x\ge0\\b=y\ge0\end{matrix}\right.\) \(\Rightarrow2y+4=\left(x+2\right)y\Rightarrow xy=4\)

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

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

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

\(P\le\dfrac{1}{2\sqrt{2}}\left(3-\dfrac{1}{x+1}+3-\dfrac{1}{y+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

\(P\le\dfrac{3\sqrt{2}}{2}-\dfrac{\sqrt{2}-1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

Ta có:

\(\dfrac{1}{x+1}+\dfrac{1}{y+1}=\dfrac{x+y+2}{xy+x+y+1}=\dfrac{x+y+2}{x+y+5}=1-\dfrac{3}{x+y+5}\ge1-\dfrac{3}{2\sqrt{xy}+5}=\dfrac{2}{3}\)

\(\Rightarrow P\le\dfrac{3\sqrt{3}}{2}-\dfrac{\sqrt{2}-1}{4}.\dfrac{2}{3}=...\)

Dấu "=" xảy ra khi \(x=y=2\) hay \(\left(a;b\right)=\left(4;2\right)\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

a.

\(F=\dfrac{a}{b+2}\Rightarrow F.b+2F=a\)

\(\Rightarrow2F=a-F.b\)

\(\Rightarrow4F^2=\left(a-F.b\right)^2\le\left(a^2+b^2\right)\left(1^2+F^2\right)=F^2+1\)

\(\Rightarrow3F^2\le1\)

\(\Rightarrow-\dfrac{1}{\sqrt{3}}\le F\le\dfrac{1}{\sqrt{3}}\)

Dấu "=" lần lượt xảy ra tại \(\left(a;b\right)=\left(-\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\) và \(\left(\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\)

b. Đặt \(\left\{{}\begin{matrix}a+b=x\\a-2b=y\end{matrix}\right.\) quay về câu a