Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\frac{4}{a+b+c}=4.\frac{4}{6}=\frac{8}{3}\)

\(\Rightarrow-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le\frac{-8}{3}\)

\(\Rightarrow M=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}\)

\(=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)

\(\Rightarrow M\le\frac{1}{3}\)

Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\a+b=c\\a+b+c=6\end{cases}\Leftrightarrow\hept{\begin{cases}a=b=\frac{3}{2}\\c=3\end{cases}}}\)

Vậy GTLN của M là 1/3

Với 2 số x,y > 0 Theo Cauchy ta có: \(\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow\frac{\left(x+y\right)^2}{4}\ge xy\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}^{\left(1\right)}\)

\(P=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}\)

\(=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Áp dụng (1) ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\cdot\frac{4}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)

\(\Rightarrow3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)

Đẳng thức xảy ra khi a=b và (a+b)=c hay a=b=1,5 và c=3.

đặt \(a+b=x,b+c=y;c+a=z\)

ta có \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\Rightarrow3-\frac{1}{x+1}-\frac{1}{y+1}-\frac{1}{z+1}=1\) \(\)

=> \(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1\)

=> \(\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{x}{x+1}=\frac{1}{x+1}\)

Áp dụng bđt cô si ta có \(\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)

=> \(\frac{1}{x+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)

tương tự ta có 

\(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\)

\(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

nhân từng vế của 3 bđt cùng chièu ta có 

\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge8\sqrt{\frac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}=8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\) 

=> \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)

\(P=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=\frac{a}{a}-\frac{1}{a}+\frac{b}{b}-\frac{1}{b}+\frac{c}{c}-\frac{4}{c}\)

=> \(P=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)(1)

Ta lại có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0< =>a+b-2\sqrt{ab}\ge0=>\frac{\left(a+b\right)^2}{4}\ge ab\)

<=> \(\frac{a+b}{ab}\ge\frac{4}{a+b}< =>\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\left(\frac{4}{a+b+c}\right)\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge4\left(\frac{4}{6}\right)=\frac{16}{6}=\frac{8}{3}\)(Do a+b+c=6 theo gt)

Thay vào (1), suy ra:

\(P=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)

=> GTLL của P là: \(P=\frac{1}{3}\)

Dấu '=' xảy ra khi a=b và a+b=c => c=3; a=b=1,5

thiếu đề bn ơi: a+b+c=?

HIHI viết thiếu nhưng mk ra rồi cảm ơn ạ !

uk, bn dùng UCT là ra mà

Cần chứng minh: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)\)

Thật vậy: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)^2\Leftrightarrow4\left(a^2-ab+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow4a^2-4ab+4b^2-a^2-b^2-2ab\ge0\Leftrightarrow3\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)

Áp dụng:\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ac+a^2}}\)

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

Dấu "=" xảy ra khi: \(a=b=c=1\)

\(b^4+c^4\ge bc\left(b^2+c^2\right)\)vì \(\left(b-c\right)^2\left(b^2+bc+c^2\right)\ge0\)

\(\Rightarrow T\le\frac{a}{\frac{b^2+c^2}{a}+a}+\frac{b}{\frac{a^2+c^2}{b}+b}+\frac{c}{\frac{a^2+b^2}{c}+c}=1\)

rõ đi bạn

Ta có:

\(P=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}\)

\(=\frac{ab}{\sqrt{1-a-b+ab}}+\frac{bc}{\sqrt{1-b-c+bc}}+\frac{ca}{\sqrt{1-a-c+ca}}\)

\(=\frac{ab}{\sqrt{\left(1-a\right)\left(1-b\right)}}+\frac{bc}{\sqrt{\left(1-b\right)\left(1-c\right)}}+\frac{ca}{\sqrt{\left(1-c\right)\left(1-a\right)}}\)

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

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

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

Vậy GTLN là  \(P=\frac{1}{2}\) khi \(a=b=c=\frac{1}{3}\)

Biến đổi một chút, ta có:\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}\)

\(=\sqrt{\frac{bc}{a+bc}}\cdot\sqrt{\frac{bc}{c+a}}\le\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\)

Tương tự cho 2 BĐT còn lại ta có: 

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

Cộng ba bất đẳng thức trên lại theo vế, ta có:

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

ai bít