\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)

TK: Cho các số thực dương a, b, c thỏa mãn a + b+ c = 3. Chứng minh rằng: \(\sqrt{2a^2+\frac{7}{b^2}}+\sqrt{2b^2+\frac{7}{... - Hoc24

Ta có : \(a+b+c=3\Rightarrow a^2+b^2+c^2\ge3\)

Theo BĐT AM - GM ta có :

\(a^4+b^2\ge2a^2b\)

\(b^4+c^2\ge2b^2c\)

\(c^4+a^2\ge2c^2a\)

\(2a^2b^2+2a^2\ge4a^2b\)

\(2b^2c^2+2b^2\ge4b^2c\)

\(2c^2a^2+2c^2\ge4c^2a\)

Cộng từng vế BĐT ta được :

\(\left(a^2+b^2+c^2\right)^2+3\left(a^2+b^2+c^2\right)\ge6\left(a^2b+b^2c+c^2a\right)\)

\(\Rightarrow a^2b+b^2c+c^2a\le\dfrac{3^2+3^2}{6}=3\)

Theo BĐT Cauchy schwarz dưới dạng en-gel ta có :

\(VT\ge\dfrac{9}{6+a^2b+b^2c+c^2a}=\dfrac{9}{9}=1\)

Dấu bằng xảy ra khi \(a=b=c=1\)

Viết lại BĐT:\(\dfrac{a^2b}{a^2b+2}+\dfrac{b^2c}{b^2c+2}+\dfrac{c^2a}{c^2a+2}\le1\)

Áp dụng BĐT AM-GM:

\(VT\le\sum\dfrac{a^2b}{3\sqrt[3]{a^4b^2}}=\dfrac{1}{3}\left(\sqrt[3]{a^2b}+\sqrt[3]{b^2c}+\sqrt[3]{c^2a}\right)\)

\(\le\dfrac{1}{9}\left(3a+3b+3c\right)=1\)

Suy ra đpcm

Ta có \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)

\(\Rightarrow\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}=\dfrac{4a^4}{a^4+2b^2a^2+a^2}\). Lập 2 BĐT tương tự rồi áp dụng bất đẳng thức BCS, ta có:

\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{\left(2a^2+2b^2+2c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\) \(=\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+3}\)\(=\dfrac{4.3^2}{3^2+3}=3\).

Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\) nên ta có đpcm. ĐTXR \(\Leftrightarrow a=b=c=1\)

\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

Tương tự ...

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

\(=\dfrac{1}{2}\left(\dfrac{c}{abc+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{ca.bc+a.bc+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c}{1+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{c+1+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c+1+bc}{1+bc+c}\right)=\dfrac{1}{2}\)

\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)

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 \(a=b=c=\dfrac{\sqrt{3}}{3}\).