1.

\(\left(1+a\right)^2=\left(1.1+\sqrt{\frac{a}{b}}.\sqrt{ab}\right)^2\le\left(1+\frac{a}{b}\right)\left(1+ab\right)=\frac{\left(a+b\right)\left(1+ab\right)}{b}\)

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

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

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

Dấu "=" xảy ra khi \(a=b=1\)

2.

\(P=\sqrt{\frac{a^2}{a^4+3}}+\sqrt{\frac{b^2}{b^4+3}}\le\sqrt{2\left(\frac{a^2}{a^4+3}+\frac{b^2}{b^4+3}\right)}\)

Đặt \(\left(a^2;b^2\right)=\left(x;y\right)\Rightarrow xy=1\)

\(Q=\frac{x}{x^2+3}+\frac{y}{y^2+3}=\frac{x}{x^2+3}+\frac{x}{3x^2+1}-\frac{1}{2}+\frac{1}{2}\)

\(Q=\frac{-\left(x-1\right)^2\left(3x^2-2x+3\right)}{2\left(x^2+3\right)\left(3x^2+1\right)}+\frac{1}{2}\le\frac{1}{2}\)

\(\Rightarrow P\le\sqrt{2Q}\le1\)

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

Bài 5.

1. Chứng minh

$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$

Ta có:

$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$

$=\dfrac{2b(a+b)+a(a+b)-4ab}{ab(a+b)}$

$=\dfrac{a^2-ab+2b^2}{ab(a+b)}$

$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$

Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$

2. Chứng minh

$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$

Vì $a,b,c>0$ nên:

$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$

Mà: $\dfrac1a>\dfrac{a}{a+b+c}$

Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$

1.

$a^3+b^4-ab(a+b)$

$=a^3+b^4-a^2b-ab^2$

$=a^2(a-b)+b^2(b-a)$

$=(a-b)(a^2-b^2)$

$=(a-b)^2(a+b)\ge0$

Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$

2.

$a^4+b^4-ab(a^2+b^2)$

$=a^4+b^4-a^3b-ab^3$

$=a^3(a-b)+b^3(b-a)$

$=(a-b)(a^3-b^3)$

$=(a-b)^2(a^2+ab+b^2)\ge0$

Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$

3.

$a^5+b^5-ab(a^3+b^3)$

$=a^5+b^5-a^4b-ab^4$

$=a^4(a-b)+b^4(b-a)$

$=(a-b)(a^4-b^4)$

$=(a-b)^2(a+b)(a^2+b^2)\ge0$

Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

Bài 2:

a) Áp dụng BĐT AM - GM ta có:

\(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}\) \(\ge2\sqrt{\dfrac{1}{4^2ab}}=\dfrac{2}{4\sqrt{ab}}=\dfrac{1}{2\sqrt{ab}}\)

\(\ge\dfrac{1}{a+b}\) (Đpcm)

b) Trừ 1 vào từng vế của BĐT ta được BĐT tương đương:

\(\left(\frac{x}{2x+y+z}-1\right)+\left(\frac{y}{x+2y+z}-1\right)+\left(\frac{z}{x+y+2z}-1\right)\le\frac{-9}{4}\)

\(\Leftrightarrow-\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le-\frac{9}{4}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)

Áp dụng BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)

\(\ge\dfrac{9}{2x+y+z+x+2y+z+x+y+2z}=\dfrac{9}{4\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)

\(\Leftrightarrow\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\le\dfrac{3}{4}\) (Đpcm)

Bài 1:

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT\ge\dfrac{\left(a+b\right)^2}{a-1+b-1}=\dfrac{\left(a+b\right)^2}{a+b-2}\)

Nên cần chứng minh \(\dfrac{\left(a+b\right)^2}{a+b-2}\ge8\)

\(\Leftrightarrow\left(a+b\right)^2\ge8\left(a+b-2\right)\)

\(\Leftrightarrow a^2+2ab+b^2\ge8a+8b-16\)

\(\Leftrightarrow\left(a+b-4\right)^2\ge0\) luôn đúng