\(3a+3b+\dfrac{1}{a+b}=\dfrac{a+b}{25}+\dfrac{1}{a+b}+\dfrac{74\left(a+b\right)}{25}\ge2.\sqrt{\dfrac{a+b}{25}.\dfrac{1}{a+b}}+\dfrac{74}{25}.5=\dfrac{76}{5}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{5}{2}\)

Vậy GTNN của biểu thức là \(\dfrac{76}{5}\)

Ta có: 3a + 3b + \(\dfrac{1}{a+b}\) = \(\dfrac{1}{a+b}+\dfrac{a+b}{25}+\dfrac{74}{25}\left(a+b\right)\)

Áp dụng BDT Co-si, ta có:

\(\dfrac{1}{a+b}+\dfrac{a+b}{25}\ge2\sqrt{\dfrac{1}{a+b}.\dfrac{a+b}{25}}\)

=> \(\dfrac{1}{a+b}+\dfrac{a+b}{25}\ge\dfrac{2}{5}\)

Mà \(\dfrac{74}{25}\left(a+b\right)\ge\dfrac{74}{5}\)

=> \(3\left(a+b\right)+\dfrac{1}{a+b}\ge\dfrac{76}{5}\)

Dấu "=" xảy ra <=> \(a=b=\dfrac{5}{2}\)

 Cosi: ab <= 1/4 
Quy đồng P, ta đc: 
P = (2ab+1)/(ab+2). 
Ta cm P <= 2/3 
<=> 3(2ab+1) <= 2(ab+2) 
<=> ab<= 1/4 (đúng) 
Vậy maxP = 2/3 khi a=b =1/2

#)Trả lời :

\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{a+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)

Tách VT = A + B và xét :

\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3b}{1+a^2}=\)\(\sum\)\(\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\)\(\sum\)\(\left(3a-\frac{3ab}{2}\right)\)

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

\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\)\(\sum\)\(ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)

( Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))

Dấu ''='' xảy ra khi a = b = c = 1

Tham khảo nhé ^^

\(4^{a+b-1}-\left(\frac{1}{2}\right)^{3a+b-2}+5a+3b-4=0\)

\(\Leftrightarrow2^{2a+2b-2}-2^{-3a-b+2}+5a+3b-4=0\)

\(\Leftrightarrow2^{2a+2b-2}+2b+2b-2=2^{-3a-b+2}-3a-b+2\)(1)

Xét hàm \(f\left(t\right)=2^t+t\)

\(f'\left(t\right)=2^t.ln\left(2\right)+1>0,\forall t\inℝ\)

suy ra \(f\left(t\right)\)đồng biến trên \(ℝ\).

(1) suy ra \(2a+2b-2=-3a-b+2\Leftrightarrow b=\frac{4-5a}{3}\)

\(P=a^2+2ab+b^2=\left(a+b\right)^2=\left(a+\frac{4-5a}{3}\right)^2\ge0\)

Dấu \(=\)khi \(a=2\).

Vậy \(minP=0\)khi \(a=2,b=-2\) 

\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)

Ta tách VT=A+B và xét

\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\text{∑}\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\text{∑}\left(3a-\frac{3ab}{2}\right)\)

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

\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\text{∑}ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)

(Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))

Dấu = khi a=b=c=1

2 + 2 =22

5555555555555 = 5 x 5 x ........

Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)

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

\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)

Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)

Đẳng thức xảy ra khi a = b = c = 1

khó phết

\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)

Ta tách VT = A + b và xét :

\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\Sigma\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\Sigma\left(3a-\frac{3ab}{2}\right)\)\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\Sigma\left(1-\frac{b^2}{1+b^2}\right)\ge\Sigma\left(1-\frac{b}{2}\right)\)

\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\Sigma ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)( Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)=3}\))

Dấu = khi a = b = c = 1 .

Yuzuri Yukari:copy câu trả lời của tôi