Áp dụng bđt Schwarz ta có:

\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).

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

Áp dụng BĐT Holder ta có: 

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

Vậy ta có thể viết lại BĐT cần chứng minh như sau;

\(VT\left(a+b\right)\le2+25ab\left(\frac{1}{\left(2a+3b\right)^2}+\frac{1}{\left(2b+3a\right)^2}\right)\)

Nó đủ để ta có thể thấy rằng 

\(25ab\left[\left(2b+3a\right)^2+\left(2a+3b\right)^2\right]\le2\left(2a+3b\right)^2\left(2b+3a\right)^2\)

\(\Leftrightarrow59\left(a^2-b^2\right)^2+13\left(a^4+b^4-a^3b-ab^3\right)\ge0\)

BĐT cuối cùng đúng nên ta có ĐPCM

ok jjj

Đặt \(\frac{a}{b}=t\)do a>0, b>0 nên t>0

Khi đó BĐT \(\frac{2a^2+3b^2}{2a^3+3b^3}+\frac{2b^2}{3b^3}+\frac{2b^2+3a^2}{2b^3+3a^2}\le\frac{4}{a+b}\left(1\right)\)trở thành

\(\frac{2t^2+3}{2t^3+3}+\frac{2+3t^2}{3+3t^3}\le\frac{4}{t+1}\)

\(\Leftrightarrow\left(2t^2+3\right)\left(2+3t^2\right)\left(t+1\right)+\left(2+3t^2\right)\left(2t^2+1\right)\left(t+1\right)\le4\left(2t^3+3\right)\left(2+3t^2\right)\)

\(\Leftrightarrow\left(t+1\right)\left(12t^5+13t^3+13t^2+12\right)\le4\left(6t^6+13t^3+6\right)\)

\(\Leftrightarrow12\left(t^6-t^5-t+1\right)-13t^2\left(t^2-12t+1\right)\ge0\)

\(\Leftrightarrow12\left(t-1\right)^2\left[12\left(t^4+t^3+t^2+t+1\right)-13t^2\right]\ge0\)

\(\Leftrightarrow\left(t-1\right)^2\left[12\left(t^4+t^3+t^2+t+1\right)-13t^2\right]\ge0\left(2\right)\)

Ta có \(12\left(t^4+t^3+t^2+t+1\right)-13t^2=12t^4+12t\left(t-1\right)^2+23t^2+12>0\forall t>0\)

BĐT (2) đúng với mọi t>0

=> BĐT (1) đúng với mọi a,b>0

Dấu "=" xảy ra <=> t=1 <=> a=b

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

\(\Leftrightarrow\frac{\left(2a^2+3b^2\right)\left(a+b\right)}{2a^3+3b^3}+\frac{\left(2b^2+3a^2\right)\left(a+b\right)}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^3+3b^3+2a^2b+3ab^2}{2a^3+3b^3}+\frac{2b^3+3a^3+2ab^2+3ab^2}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^2b+3ab^2}{2a^3+3b^3}+\frac{2ab^2+3ab^2}{2b^3+3a^3}\le2\)

\(\Leftrightarrow\frac{2\left(\frac{a}{b}\right)^2+3\left(\frac{a}{b}\right)}{2\left(\frac{a}{b}\right)^3+3}+\frac{2\left(\frac{a}{b}\right)+3\left(\frac{a}{b}\right)^2}{3\left(\frac{a}{b}\right)^3+2}\le2\)

Đặt \(\frac{a}{b}=x>0\Rightarrow\frac{2x^2+3x}{2x^3+3}+\frac{3x^2+2x}{3x^3+2}\le2\)

\(\Leftrightarrow\left(x-1\right)^2\left(12x^4+12x^3-x^2+12x+12\right)\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=1\) hay \(a=b\)

Hơi trâu bò :D

quy đồng mẫu số ta được

\(\frac{\left(a-b\right)^2}{a\left(a^2-b^2\right)}+\frac{\left(a+b\right)^2}{a\left(a^2-b^2\right)}=\frac{a\left(3a-b\right)}{a\left(a^2-b^2\right)}\)<=> (a-b)² +(a+b)² = a(3a-b) <=> a²- ab- 2b²= 0 <=> (a+ b)(a- 2b) = 0

<=> a=-b hoăc a =2b

với a= -b => P= \(\frac{-b^3+2b^3+2b^3}{-2b^3-b^3+2b^3}=-3\)

với a =2b => P= \(\frac{\left(2b\right)^3+2.\left(2b\right)^2b+2b^3}{2.\left(2b\right)^3+2b.b^2+2b^3}=\frac{3}{2}\)

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