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3)

1.

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)

Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)

\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)

\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)

\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)

\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)

2.

Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)

Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)

\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)

Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)

\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)

\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)

(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị

Bài 1:

Áp dụng BĐT Bunhiacopxky ta có:

$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$

$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$

$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)

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

Bài 2: 

Áp dụng BĐT Bunhiacopxky:

$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$

$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$

$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$

$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>

Bài 3:

Áp dụng BĐT Bunhiacopxky:

$2=(a^2+b^2)(1+1)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2}$

$(a\sqrt{1+a}+b\sqrt{1+b})^2\leq (a^2+b^2)(1+a+1+b)$

$=2+a+b\leq 2+\sqrt{2}$

$\Rightarrow a\sqrt{1+a}+b\sqrt{1+b}\leq \sqrt{2+\sqrt{2}}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=\frac{1}{\sqrt{2}}$

3/ Áp dụng bất đẳng thức AM-GM, ta có :

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

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

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

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

1.

BĐT cần chứng minh tương đương:

\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)

Ta có:

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

\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)

Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)

\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)

Do \(a;b;c\ge1\)  nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:

\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)

\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Câu 2 em kiểm tra lại đề có chính xác chưa

2.

Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS

Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)

\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)

BĐT cần chứng minh tương đương:

\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

Đúng theo (1)

Dấu "=" xảy ra khi \(a=b=c\)

\(VT=\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}-3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)

-Áp dụng BĐT Caushy Schwarz cho 3 số dương ta có:

\(VT\ge\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{a+b+b+c+c+a}-3=\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}-3=\dfrac{9}{2}-3=\dfrac{3}{2}\left(1\right)\)

\(VP=\dfrac{2.\left(\dfrac{a}{a^2+1}+\dfrac{1}{2}+\dfrac{b}{b^2+1}+\dfrac{1}{2}+\dfrac{c}{c^2+1}+\dfrac{1}{2}-\dfrac{3}{2}\right)}{2}=\dfrac{\dfrac{2a}{a^2+1}+1+\dfrac{2b}{b^2+1}+1+\dfrac{c}{c^2+1}-3}{2}=\dfrac{\dfrac{a^2+2a+1}{a^2+1}+\dfrac{b^2+2b+1}{b^2+1}+\dfrac{c^2+2c+1}{c^2+1}-3}{2}=\dfrac{\dfrac{\left(a+1\right)^2}{a^2+1}+\dfrac{\left(b+1\right)^2}{b^2+1}+\dfrac{\left(c+1\right)^2}{c^2+1}-3}{2}\)-Áp dụng BĐT Caushy ta có:

\(VP\le\dfrac{\dfrac{2\left(a^2+1\right)}{a^2+1}+\dfrac{2\left(b^2+1\right)}{b^2+1}+\dfrac{2\left(c^2+1\right)}{c^2+1}-3}{2}=\dfrac{2+2+2-3}{2}=\dfrac{3}{2}\left(2\right)\)

-Từ (1) và (2) ta có:

\(VT\ge\dfrac{3}{2}\ge VP\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\ge\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\left(đpcm\right)\)

-Dấu bằng xảy ra \(\Leftrightarrow a=b=c=1\)

\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\\ =\left(ab^2-a\right)\left(c^2-1\right)+\left(a^2b-b\right)\left(c^2-1\right)+\left(a^2c-c\right)\left(b^2-1\right)\\ =ab^2c^2-ab^2-ac^2+a+a^2bc^2-a^2b-bc^2+b+a^2b^2c-a^2c-b^2c+c\\ =abc\left(ab+bc+ac\right)-\left(a^2b+ab^2+ac^2+bc^2+a^2c+b^2c\right)+\left(a+b+c\right)\\ =abc\left(ab+bc+ca\right)+\left(a+b+c\right)+3abc-\left[\left(a^2b+ab^2+abc\right)+\left(b^2c+bc^2+abc\right)+\left(a^2c+ac^2+abc\right)\right]\\ =abc\left(ab+bc+ca\right)+abc+3abc-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\right]\\ =4abc+abc\left(ab+bc+ca\right)-\left(a+b+c\right)\left(ab+bc+ca\right)\\ =4abc+abc\left(ab+bc+ca\right)-abc\left(ab+bc+ca\right)=4abc\)