Với các số dương x;y ta có:

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

Áp dụng:

\(\Rightarrow P=\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}+\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{a}{ca\left(c+a\right)+abc}\)

\(\Rightarrow P\le\dfrac{abc}{ab\left(a+b+c\right)}+\dfrac{abc}{bc\left(a+b+c\right)}+\dfrac{abc}{ca\left(a+b+c\right)}\)

\(\Rightarrow P\le\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(P_{max}=1\) khi \(a=b=c=1\)

\(BDT\Leftrightarrow\frac{a^3}{\left(1-a\right)^2}+\frac{b^3}{\left(1-b\right)^2}+\frac{c^3}{\left(1-c\right)^2}\ge\frac{1}{4}\)

Ta có BĐT phụ: \(\frac{a^3}{\left(1-a\right)^2}\ge a-\frac{1}{4}\)

\(\Leftrightarrow\frac{\left(3a-1\right)^2}{4\left(a-1\right)^2}\ge0\forall0< a\le\frac{1}{3}\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\frac{b^3}{\left(1-b\right)^2}\ge b-\frac{1}{4};\frac{c^3}{\left(1-c\right)^2}\ge c-\frac{1}{4}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\left(a+b+c\right)-\frac{1}{4}\cdot3=1-\frac{3}{4}=\frac{1}{4}=VP\)

Xảy ra khi \(a=b=c=\frac{1}{3}\)

Áp dụng BĐT cô si ta có:

\(\frac{a^3}{\left(b+c\right)^2}+\frac{1a}{4}\ge\frac{a^2}{b+c}\)\(,\frac{b^3}{\left(c+a\right)^2}+\frac{1b}{4}\ge\frac{b^2}{a+c},\frac{c^3}{\left(a+b\right)^2}+\frac{1c}{4}\ge\frac{c^2}{a+b}\)

Cộng lại ta có

\(VT\ge\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}-\frac{1}{4}\left(a+b+c\right)\)

\(\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}-\frac{1}{4}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\left(đpcm\right)\)

Dấu =tự tìm Ok

thanks cả 2 nhé

Với a, b, c là các số dương.

Ta có: \(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}=2\)

\(\Rightarrow\) \(\dfrac{1}{a+b+1}=\left(1-\dfrac{1}{b+c+1}\right)+\left(1-\dfrac{1}{c+a+1}\right) \)

\(=\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}\)

\(\ge2\sqrt{\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}>0\) (Bất đẳng thức Cô-si)

Tương tự: \(\dfrac{1}{b+c+1}\ge2\sqrt{\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c+a+1\right)\left(a+b+1\right)}}>0\)

\(\dfrac{1}{c+a+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}>0\)

Nhân vế theo vế ba bất đẳng thức trên, ta được:

\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\) \(\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)

\(\Rightarrow\) \(1\ge8\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Rightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\dfrac{1}{4}\).

Vậy giá trị lớn nhất của tích \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\) bằng \(\dfrac{1}{8}\) khi và chỉ khi \(a=b=c=\dfrac{1}{4}\).

\(P\ge\frac{4}{a+b}-c=1+\frac{c+1}{3-c}-c=1+\frac{c^2-2a+1}{3-c}=1+\frac{\left(c-1\right)^2}{3-c}\ge1\) ; \(\forall c< 3\)

\(\Rightarrow P_{min}=1\) khi \(a=b=c=1\)

Ta có:

\(\sqrt[4]{4}VT=\sqrt[4]{4a^3}+\sqrt[4]{4b^3}+\sqrt[4]{4c^3}\)

\(=\sqrt[4]{\left(a+b+c\right)a^3}+\sqrt[4]{\left(a+b+c\right)b^3}+\sqrt[4]{\left(a+b+c\right)c^3}\)

\(>\sqrt[4]{a^4}+\sqrt[4]{b^4}+\sqrt[4]{c^4}=a+b+c=4\)

\(\Rightarrow\sqrt[4]{4}VT>4\Rightarrow VT>\dfrac{4}{\sqrt[4]{4}}=2\sqrt{2}\)

Ta có: \(\frac{19a+3}{b^2+1}=\left(19a+3\right).\frac{1}{b^2+1}=\left(19a+3\right)\left(1-\frac{b^2}{b^2+1}\right)\)

\(\ge\left(19a+3\right)\left(1-\frac{b^2}{2b}\right)=\left(19a+3\right)\left(1-\frac{b}{2}\right)\)

\(=19a+3-\frac{19ab}{2}-\frac{3b}{2}\)(1)

Hoàn toàn tương tự, ta có: \(\frac{19b+3}{c^2+1}\ge19b+3-\frac{19bc}{2}-\frac{3c}{2}\)(2); \(\frac{19c+3}{a^2+1}\ge19c+3-\frac{19ca}{2}-\frac{3a}{2}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(A=\frac{19a+3}{b^2+1}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)\(\ge19\left(a+b+c\right)-\frac{3\left(a+b+c\right)}{2}-\frac{19\left(ab+bc+ca\right)}{2}+9\)

\(=\frac{35\left(a+b+c\right)}{2}-\frac{19\left(ab+bc+ca\right)}{2}+9\)

\(\ge\frac{35.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{19.3}{2}+9=\frac{105}{2}-\frac{57}{2}+9=33\)

Đẳng thức xảy ra khi a = b = c = 1.