\(a+b+c=2\Rightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{4}{3}\)

\(P=\dfrac{7+2b}{1+a}+\dfrac{7+2c}{1+b}+\dfrac{7+2a}{1+c}\)

\(\ge\dfrac{\left(21+2\left(a+b+c\right)\right)^2}{\left(1+a\right)\left(7+2b\right)+\left(1+b\right)\left(7+2c\right)+\left(1+c\right)\left(7+2a\right)}\)

\(=\dfrac{25^2}{21+9\left(a+b+c\right)+2\left(ab+bc+ca\right)}\ge\dfrac{25^2}{21+9.2+\dfrac{2.4}{3}}=15\)

\("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

Bài 1:

Ta có:

\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)

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

\(=3-2M(*)\)

Áp dụng BĐT Cauchy ta có:

\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)

\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)

Tiếp tục áp dụng BĐT Cauchy:

\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)

Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)

\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)

Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)

(đpcm)

Dấu bằng xảy ra khi $a=b=c=1$

Bài 2:

Áp dụng BĐT Cauchy -Schwarz:

\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)

hay:

\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)

Mặt khác, theo BĐT Cauchy ta dễ thấy:

\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)

\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)

\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)

Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)

Ta có đpcm

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

1) Ta c/m BĐT sau:

Với a, b > 0 thì \(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)

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

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì a, b > 0)

Đẳng thức xảy ra \(\Leftrightarrow a=b\)

Như vậy ta có \(\left\{{}\begin{matrix}x^3+y^3\ge xy\left(x+y\right)\\y^3+z^3\ge yz\left(y+z\right)\\z^3+x^3\ge zx\left(z+x\right)\end{matrix}\right.\)

Do đó \(VT\ge\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}+\dfrac{\sqrt{xyz+yz\left(y+z\right)}}{yz}+\dfrac{\sqrt{xyz+zx\left(z+x\right)}}{zx}\)

\(=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}+\dfrac{\sqrt{yz\left(x+y+z\right)}}{yz}+\dfrac{\sqrt{zx\left(x+y+z\right)}}{zx}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(=\sqrt{x+y+z}.\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\)

\(=\sqrt{x+y+z}.\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\ge\sqrt{3\sqrt[3]{xyz}}.3\sqrt[3]{\sqrt{xyz}}=3\sqrt{3}\)

Đẳng thức xảy ra \(\Leftrightarrow x=y=z=1\)

1) Lợi dụng BĐT AM-GM cho 3 số dương, ta được:

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3.y^3.1}}}{xy}=\sqrt{\dfrac{3}{xy}}\)

Tương tự:

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}\ge\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+x^3+z^3}}{xz}\ge\sqrt{\dfrac{3}{xz}}\)

Cộng từng vế các BĐT trên. ta được:

\(VT\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

Tiếp tục lợi dụng AM-GM, ta được

\(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\ge3\sqrt[3]{\dfrac{1}{\sqrt{xy}}.\dfrac{1}{\sqrt{yz}}.\dfrac{1}{\sqrt{xz}}}=3\)

Suy ra đpcm. Đẳng thức xảy ra khi x=y=z=1

Ta có BĐT sau:\(\dfrac{1}{1-a^2}+\dfrac{1}{1-b^2}\ge\dfrac{2}{1-ab}\left(\forall a,b\in\left(0;1\right)\right)\)(*)

Cm:(*)\(\Leftrightarrow\dfrac{\left(ab+1\right)\left(a-b\right)^2}{\left(1-a^2\right)\left(1-b^2\right)\left(1-ab\right)}\ge0\)( đúng vì 0<a,b<1)

\(VT=\dfrac{1}{2}\left[\sum\dfrac{2a^2}{1-a^2}\right]=\dfrac{1}{2}\left[\sum\left(\dfrac{2a^2}{1-a^2}+2\right)\right]-3\)

\(=\dfrac{1}{2}\left[\sum\left(\dfrac{2}{1-a^2}\right)\right]-3=\dfrac{1}{2}\sum\left(\dfrac{1}{1-a^2}+\dfrac{1}{1-b^2}\right)-3\ge\dfrac{1}{2}.\sum\dfrac{2}{1-ab}-3=1\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)

a)Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+1\ge2y\end{matrix}\right.\)\(\Rightarrow x^2+2y^2+1\ge2xy+2y\)

\(\Rightarrow x^2+2y^2+3\ge2xy+2y+2\)

\(\Rightarrow\dfrac{1}{x^2+2y^2+3}\le\dfrac{1}{2\left(xy+y+1\right)}\Leftrightarrow\dfrac{2}{x^2+2y^2+3}\le\dfrac{1}{xy+y+1}\)

b)Áp dụng bổ đề trên ta có:

\(a^2+2b^2+3\ge2ab+2b+2\Rightarrow\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2\left(ab+b+1\right)}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2\left(bc+b+1\right)};\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2\left(ac+c+1\right)}\)

Cộng theo vế 3 BĐT trên ta có:

\(Q\le\dfrac{1}{2\left(ab+b+1\right)}+\dfrac{1}{2\left(bc+b+1\right)}+\dfrac{1}{2\left(ac+c+1\right)}\)

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

\(=\dfrac{1}{2}\left(\dfrac{a}{ac+c+1}+\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}\right)\left(abc=1\right)\)

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

Đẳng thức xảy ra khi \(x=y=z=1\)

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a+b+c=5

=>a+b=5-c

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

=>\(ab + b \le \frac{(5 - c + 1)^2}{4} = \frac{(6 - c)^2}{4}\)

a,b,c>0

=>ab+b>0

=>\(\frac{c^3}{ab + b} \ge \frac{c^3}{\frac{(6 - c)^2}{4}} = \frac{4c^3}{(6 - c)^2}\)

4a+4b=4(a+b)=4(5-c)=20-4c

=>P>=20-4c+\(\frac{4c^3}{\left(6-c\right)^2}\)

\(\frac{4c^3}{(6 - c)^2} = \frac{4c^3}{(6 - c)^2} + (6 - c) + (6 - c) - 2(6 - c)\)

Theo Cosi, ta sẽ có:

\(\frac{4c^3}{(6 - c)^2} + (6 - c) + (6 - c) \ge 3 \sqrt[3]{\frac{4c^3}{(6 - c)^2} \cdot (6 - c) \cdot (6 - c)} = 3 \sqrt[3]{4c^3} = 3c \sqrt[3]{4}\)

\(\frac{4c^3}{(6 - c)^2}-\frac{25}{12}c+\frac{125}{36}\)

\(=\frac{144c^3 - (75c - 125)(36 - 12c + c^2)}{36(6 - c)^2}\)

\(= \frac{144c^3 - (75c^3 - 900c^2 + 2700c - 125c^2 + 1500c - 4500)}{36(6 - c)^2}\)

\(= \frac{69c^3 + 1025c^2 - 4200c + 4500}{36(6 - c)^2}\)

\(= \frac{(3c - 5)^2 (23c + 180)}{36(6 - c)^2} \ge 0 \quad \forall c \in (0, 5)\)

=>\(\frac{4c^3}{(6 - c)^2} \ge \frac{25}{12}c - \frac{125}{36}\) ∀c∈(0;5)

Dấu '=' xảy ra khi 3c-5=0

=>c=5/3

Do đó, ta có: \(P \ge 20 - 4c + \frac{25}{12}c - \frac{125}{36}\)

=>\(P \ge 20 - \frac{23}{12}c - \frac{125}{36}\)

=>\(P \ge 20 - 4\left(\frac{5}{3}\right) + \frac{4\left(\frac{5}{3}\right)^3}{\left(6 - \frac{5}{3}\right)^2}\)

=>\(P \ge 20 - \frac{20}{3} + \frac{4 \cdot \frac{125}{27}}{\frac{169}{9}} = \frac{40}{3} + \frac{500}{507} = \frac{7260}{507} = \frac{2420}{169}\)

Vậy: GTNN của P là 2420/169 khi a=b=c=5/3