Áp dụng bất đẳng thức Cauchy ta có :

\(\frac{a}{1+b}+\frac{4}{9}.a\left(1+b\right)\ge2\sqrt{\frac{a.4.a.\left(1+b\right)}{\left(1+b\right)9}}=2\sqrt{\frac{4a^2}{3^2}}=\frac{4a}{3}\)

\(\frac{b}{1+a}+\frac{4}{9}.b\left(1+a\right)\ge2\sqrt{\frac{b.4.b.\left(1+a\right)}{\left(1+a\right)9}}=2\sqrt{\frac{2^2b^2}{3^2}}=\frac{4b}{3}\)

Cộng theo vế các bất đẳng thức cùng chiều ta được :

\(\frac{a}{1+b}+\frac{b}{1+a}+\frac{4}{9}.a\left(1+b\right)+\frac{4}{9}.b\left(1+a\right)\ge\frac{4a}{3}+\frac{4b}{3}\)

\(< =>\frac{a}{1+b}+\frac{b}{1+a}\ge\frac{4a}{3}-\frac{4}{9}\left(a+ab\right)-\frac{4}{9}\left(b+ab\right)+\frac{4b}{3}\)

\(< =>\frac{a}{1+b}+\frac{b}{1+a}\ge\frac{8a}{9}+\frac{8b}{9}-\frac{4}{9}ab-\frac{4}{9}ab\)

\(< =>S\ge\frac{1}{a+b}+\frac{8}{9}\left(a+b\right)-\frac{8}{9}ab=\left(\frac{1}{a+b}+a+b\right)-\frac{a+b+8ab}{9}\)

\(< =>S\ge2-\frac{a+b+8ab}{9}\)

Do \(4ab\le\left(a+b\right)^2\le1< =>a+b+8ab\le3\)

Khi đó ta được : \(S\ge2-\frac{3}{9}=2-\frac{1}{3}=\frac{5}{3}\).Đẳng thức xảy ra \(< =>a=b=\frac{1}{2}\)

Vậy GTNN của \(S=\frac{5}{3}\)đạt được khi \(a=b=\frac{1}{2}\)

Đầu tiên,ta chứng minh BĐT phụ \(\frac{\left(x+y\right)^2}{2}\ge2xy\Leftrightarrow\frac{\left(x+y\right)^2-4xy}{2}\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng).Dấu "=" xảy ra khi x = y.

Và BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\).Áp dụng BĐT AM-GM(Cô si),ta có; \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{\left(x+y\right)}{2}}=\frac{4}{x+y}\)

Dấu "=" xảy ra khi x = y

\(P=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\)\(\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2ab}\)

\(\ge\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}\ge4+\frac{1}{\frac{1}{2}}=6\)

Dấu "=" xảy ra khi a = b và a + b = 1 tức là a=b=1/2

Vậy Min P = 6 khi a = b = 1/2 

\(A=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\ge\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)}+\frac{4}{ab\left(a+b\right)}\)

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

\(\ge\frac{\left(1+1+1+1\right)^2}{\left(a+b\right)^2}+\frac{1}{ab}\ge\frac{16}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{4}}\ge16+4=20\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{2}\)

Ta dễ có:

\(2+4ab=\left(a+b\right)^2+a+b\ge4ab+a+b\Rightarrow a+b\le2\)

\(P=\frac{a^2-2a+2}{b+1}+\frac{b^2-2b+2}{a+1}\)

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

\(\ge\frac{\left(a+b-2\right)^2}{a+b+2}+\frac{4}{a+b+2}\ge\frac{\left(a+b-2\right)^2}{a+b+2}+1\ge1\)

Đẳng thức xảy ra tại \(a=b=1\)

hmm check hộ mình nhá

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

\(=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(\frac{3}{2ab}+24ab\right)-20ab\)

\(\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{3}{2ab}.24ab}-\frac{20\left(a+b\right)^2}{4}\ge11\) (sử dụng BĐT Cô si và giả thiết \(a+b\le1\))

Đẳng thức xảy ra khi \(a=b=\frac{1}{2}\)

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

\(P=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}\)

\(P=\frac{1}{a^2+b^2+1}+\frac{\frac{1}{9}}{2ab}+\frac{4}{9ab}\)

\(\ge\frac{\left(1+\frac{1}{3}\right)^2}{a^2+b^2+1+2ab}+\frac{4}{9ab}\)

\(\ge\frac{\left(1+\frac{3}{4}\right)^2}{\left(a+b\right)^2+1}+\frac{16}{9\left(a+b\right)^2}\)

\(\ge\frac{\left(1+\frac{1}{3}\right)^2}{1+1}+\frac{16}{9}=\frac{8}{3}\)

Dấu = xảy ra khi \(a=b=\frac{1}{2}\)