a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

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

\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)

Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)

Cộng vế với vế:

\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Áp dụng BĐT Cosi ta có \(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\ge2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)

Tương tự \(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc}\ge1\) \(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ca}\ge1\)

Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được

\(\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\right)\ge\frac{3}{4}\)

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

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

\(\Leftrightarrow\frac{3}{4}\ge\frac{3}{4}\) luôn đúng. Từ đó suy ba BĐT được chứng minh. Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Đặt \(\frac{a}{b}=t\)

Ta có:\(t^2+\frac{1}{t^2}+4\ge3\left(t+\frac{1}{t}\right)\)

\(\Leftrightarrow t^2+\frac{1}{t^2}+4-3t-\frac{3}{t}\ge0\)

\(\Leftrightarrow\left(t^2-2t+1\right)+\left(\frac{1}{t^2}-\frac{3}{t}+1\right)+2-t-\frac{1}{t}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2+\left(\frac{1}{t}-1\right)^2+1-t-\frac{1}{t}+t\cdot\frac{1}{t}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2+\left(\frac{1}{t}-1\right)^2+\left(t-1\right)\left(\frac{1}{t}-1\right)\ge0\)

Đặt \(\left(t-1;\frac{1}{t}-1\right)\rightarrow\left(p,q\right)\)

Ta có:

\(p^2+q^2+pq\ge0\)

\(\Leftrightarrow\left(p^2+pq+\frac{q^2}{4}\right)+\frac{3q^2}{4}\ge0\)

\(\Leftrightarrow\left(p+\frac{q}{2}\right)^2+\frac{3q^2}{4}\ge0\) *luôn đúng*

Lời giải :

Đặt \(\frac{a}{b}=t\Leftrightarrow\frac{b}{a}=\frac{1}{t}\)

BĐT \(\Leftrightarrow t^2+\frac{1}{t^2}+4\ge3\left(t+\frac{1}{t}\right)\)

\(\Leftrightarrow\left(t+\frac{1}{t}\right)^2-3\left(t+\frac{1}{t}\right)+2\ge0\)

\(\Leftrightarrow\left(t+\frac{1}{t}-1\right)\left(t+\frac{1}{t}-2\right)\ge0\)

\(\Leftrightarrow\frac{t^2-t+1}{t}\cdot\frac{t^2-2t+1}{t}\ge0\)

\(\Leftrightarrow\frac{\left(t^2-t+1\right)\left(t-1\right)^2}{t^2}\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow t=1\Leftrightarrow\frac{a}{b}=1\Leftrightarrow a=b\)

\(\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}=\frac{a^6}{a^2b+a^2c}+\frac{b^6}{b^2a+b^2c}+\frac{c^6}{c^2a+c^2b}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)}=\frac{a^3+b^3+c^3}{2}\)